1. Solving General Trigonometric Equation

Solving General Trigonometric Equations

Trigonometric equations are equations involving sine, cosine, tangent, or other trigonometric functions. Unlike many algebraic equations, trigonometric equations often have infinitely many solutions because trigonometric functions are periodic.

For example:

sin x = 1/2

has solutions at 30° and 150°, but these are not the only solutions. Because sine repeats every 360°, additional solutions occur at 390°, 510°, 750°, and so on.

To solve trigonometric equations successfully, we must understand how to find solutions within a specified interval, express all solutions using general formulas, and verify that the answers satisfy the original equation.

More advanced problems may involve multiple angles, compound angles, quadratic expressions, identities, or combinations of different trigonometric functions.

Understanding Trigonometric Equations

A trigonometric equation contains one or more trigonometric functions of an unknown angle.

Examples include:

• sin x = 1/2

• cos x = −√3/2

• tan x = 1

• 2sin x − 1 = 0

• sin 2x = cos x

• 2cos² x − 3cos x + 1 = 0

The goal is to determine every angle that makes the equation true within the required domain or interval.

Equations Versus Identities

A trigonometric equation is true only for particular values of the variable.

For example:

sin x = 1/2

is true at certain angles.

A trigonometric identity is true for every angle in its domain.

For example:

sin² x + cos² x = 1

This distinction is important because identities are used to transform equations into forms that are easier to solve.

Periodicity and Multiple Solutions

Trigonometric functions repeat their values at regular intervals.

The length of one complete cycle is called the period.

For example:

sin 30° = 1/2

sin 390° = 1/2

sin 750° = 1/2

Each angle differs from the previous one by 360°.

Similarly:

tan 45° = 1

tan 225° = 1

tan 405° = 1

Tangent repeats every 180°.

General Principle

If x is a solution to a sine or cosine equation, then adding or subtracting a whole number of complete periods produces another solution.

For tangent, the repeated interval is 180° rather than 360°.

This explains why general trigonometric equations often have infinitely many solutions.

Solving Equations over Specified Intervals

A specified interval limits the possible solutions.

For example:

0° ≤ x < 360°

means that x may equal 0° but must be less than 360°.

Another common interval is:

0 ≤ x < 2π

which represents one complete revolution in radians.

The endpoints matter.

For example, in:

0° ≤ x < 360°

the angle 0° is included, but 360° is excluded.

In:

0° ≤ x ≤ 360°

both endpoints are included.

Even though 0° and 360° are coterminal, they are distinct numerical values and may both need to be listed when the interval includes them.

Reference Angles and Quadrants

A reference angle is the acute angle formed between an angle's terminal side and the x-axis.

Reference angles help determine all solutions to a trigonometric equation.

For example:

sin x = 1/2

The reference angle is:

30°

Sine is positive in Quadrants I and II.

Therefore:

x = 30°

or:

x = 180° − 30° = 150°

Within one complete revolution:

x = 30°, 150°

Signs of Trigonometric Functions

A useful memory aid is ASTC:

• Quadrant I: All positive.

• Quadrant II: Sine positive.

• Quadrant III: Tangent positive.

• Quadrant IV: Cosine positive.

These signs help identify which quadrants contain valid solutions.

Solving Basic Sine Equations

Worked Example

Solve:

sin x = √3/2

for:

0° ≤ x < 360°

Find the Reference Angle

Since:

sin 60° = √3/2

the reference angle is 60°.

Determine the Quadrants

Sine is positive in Quadrants I and II.

Quadrant I:

x = 60°

Quadrant II:

x = 180° − 60° = 120°

Final Answer

x = 60°, 120°

Verification

sin 60° = √3/2

sin 120° = √3/2

Both solutions satisfy the original equation.

Solving Negative Sine Equations

Worked Example

Solve:

sin x = −1/2

for:

0° ≤ x < 360°

The reference angle is:

30°

Sine is negative in Quadrants III and IV.

Quadrant III:

x = 180° + 30° = 210°

Quadrant IV:

x = 360° − 30° = 330°

Therefore:

x = 210°, 330°

Solving Basic Cosine Equations

Worked Example

Solve:

cos x = −√2/2

for:

0° ≤ x < 360°

The reference angle is:

45°

Cosine is negative in Quadrants II and III.

Quadrant II:

x = 180° − 45° = 135°

Quadrant III:

x = 180° + 45° = 225°

Therefore:

x = 135°, 225°

Verification

cos 135° = −√2/2

cos 225° = −√2/2

Both solutions are valid.

Solving Basic Tangent Equations

Worked Example

Solve:

tan x = √3

for:

0° ≤ x < 360°

The reference angle is:

60°

Tangent is positive in Quadrants I and III.

Quadrant I:

x = 60°

Quadrant III:

x = 180° + 60° = 240°

Therefore:

x = 60°, 240°

Important Difference

Tangent has a period of 180°.

Therefore, its solutions can be written using one repeating family rather than two.

Determining General Solutions

A general solution describes all possible solutions to a trigonometric equation.

The integer n represents the number of complete periods added or subtracted.

n ∈ ℤ

This means n may be any integer:

…, −2, −1, 0, 1, 2, …

General Solutions for Sine

For:

sin x = a

where −1 ≤ a ≤ 1, let:

α = arcsin(a)

Then the general solutions in radians are:

x = α + 2πn

or:

x = π − α + 2πn

where n ∈ ℤ.

In degrees:

x = α + 360°n

or:

x = 180° − α + 360°n

where α is measured in degrees.

General Solutions for Cosine

For:

cos x = a

where −1 ≤ a ≤ 1, let:

α = arccos(a)

Then:

x = ±α + 2πn

or, in degrees:

x = ±α + 360°n

where n ∈ ℤ.

General Solutions for Tangent

For:

tan x = a

let:

α = arctan(a)

Then:

x = α + πn

or, in degrees:

x = α + 180°n

where n ∈ ℤ.

Worked Example: General Sine Solutions

Find the general solution of:

sin x = 1/2

The reference angle is:

30°

The solutions within one revolution are:

30° and 150°.

Since sine repeats every 360°:

x = 30° + 360°n

or:

x = 150° + 360°n

where n ∈ ℤ.

Interpreting the Integer n

If n = 0:

x = 30°, 150°

If n = 1:

x = 390°, 510°

If n = −1:

x = −330°, −210°

All these angles satisfy the equation.

Worked Example: General Cosine Solutions

Find the general solution of:

cos x = −1/2

Cosine equals −1/2 at:

120° and 240°.

Therefore:

x = 120° + 360°n

or:

x = 240° + 360°n

where n ∈ ℤ.

An equivalent compact form is:

x = ±120° + 360°n

because −120° is coterminal with 240°.

Worked Example: General Tangent Solutions

Find the general solution of:

tan x = −1

The reference angle is:

45°

Tangent is negative in Quadrants II and IV.

Within one revolution:

x = 135°, 315°

Since tangent repeats every 180°:

x = 135° + 180°n

where n ∈ ℤ.

An equivalent general solution is:

x = −45° + 180°n

Both describe the same infinite set of angles.

Solving Trigonometric Equations in Radians

The methods are identical when angles are measured in radians.

The important special-angle values are:

Worked Example

Solve:

sin x = √2/2

for:

0 ≤ x < 2π

The reference angle is:

π/4

Sine is positive in Quadrants I and II.

Therefore:

x = π/4, 3π/4

General Solution

x = π/4 + 2πn

or:

x = 3π/4 + 2πn

where n ∈ ℤ.

Solving Equations Requiring Rearrangement

Many trigonometric equations must first be rearranged into a standard form.

Worked Example

Solve:

3sin x − 1 = 0

for:

0° ≤ x < 360°

Rearrange:

3sin x = 1

sin x = 1/3

Find the reference angle:

α = arcsin(1/3)

α ≈ 19.47°

Sine is positive in Quadrants I and II.

Quadrant I:

x ≈ 19.47°

Quadrant II:

x ≈ 180° − 19.47°

x ≈ 160.53°

Final Answer

x ≈ 19.47°, 160.53°

General Solution

x ≈ 19.47° + 360°n

or:

x ≈ 160.53° + 360°n

where n ∈ ℤ.

For exact general solutions, retain arcsin(1/3) rather than rounding the reference angle.

Recognising Equations with No Real Solutions

Sine and cosine have a range of:

−1 ≤ sin x ≤ 1

−1 ≤ cos x ≤ 1

Therefore:

sin x = 2

has no real solutions.

Similarly:

cos x = −1.5

has no real solutions.

Worked Example

Solve:

4cos x + 5 = 0

Rearrange:

4cos x = −5

cos x = −5/4

Since:

−5/4 = −1.25

is outside the range of cosine:

There are no real solutions.

Recognising the range can prevent unnecessary calculations.

Solving Equations Involving Multiple Angles

Expressions such as sin 2x and cos 3x complete more cycles than sin x and cos x over the same interval.

This means they may have more solutions.

Worked Example

Solve:

sin 2x = 1/2

for:

0° ≤ x < 360°

Let:

u = 2x

Then:

0° ≤ u < 720°

Solve:

sin u = 1/2

Within this expanded interval:

u = 30°, 150°, 390°, 510°

Since:

u = 2x

divide each solution by 2:

x = 15°, 75°, 195°, 255°

Final Answer

x = 15°, 75°, 195°, 255°

General Solution

From:

2x = 30° + 360°n

or:

2x = 150° + 360°n

we obtain:

x = 15° + 180°n

or:

x = 75° + 180°n

where n ∈ ℤ.

Solving Triple-Angle Equations

Worked Example

Solve:

cos 3x = 0

for:

0° ≤ x < 360°

Let:

u = 3x

Then:

0° ≤ u < 1080°

Cosine equals zero at:

u = 90°, 270°, 450°, 630°, 810°, 990°

Divide each value by 3:

x = 30°, 90°, 150°, 210°, 270°, 330°

Final Answer

x = 30°, 90°, 150°, 210°, 270°, 330°

General Solution

Cosine equals zero when:

3x = 90° + 180°n

Therefore:

x = 30° + 60°n

where n ∈ ℤ.

This single expression describes all solutions.

Solving Compound-Angle Equations

Compound-angle equations contain expressions such as:

sin(x + 30°)

cos(x − 45°)

tan(2x + 15°)

The main strategy is to isolate the complete angle expression before solving for x.

Worked Example

Solve:

sin(x − 30°) = 1/2

for:

0° ≤ x < 360°

Let:

u = x − 30°

The general solutions are:

u = 30° + 360°n

or:

u = 150° + 360°n

Substitute back:

x − 30° = 30° + 360°n

or:

x − 30° = 150° + 360°n

Solve:

x = 60° + 360°n

or:

x = 180° + 360°n

Within the specified interval:

x = 60°, 180°

Solving Equations with Both Multiple Angles and Phase Shifts

Worked Example

Solve:

cos(2x − 30°) = 1/2

for:

0° ≤ x < 360°

Let:

u = 2x − 30°

The general solutions of cos u = 1/2 are:

u = 60° + 360°n

or:

u = −60° + 360°n

First Family

2x − 30° = 60° + 360°n

2x = 90° + 360°n

x = 45° + 180°n

Within the interval:

x = 45°, 225°

Second Family

2x − 30° = −60° + 360°n

2x = −30° + 360°n

x = −15° + 180°n

Within the interval:

x = 165°, 345°

Final Answer

x = 45°, 165°, 225°, 345°

Solving Quadratic Trigonometric Equations

Some equations contain squared trigonometric functions and can be solved using algebraic factorisation.

Worked Example

Solve:

2sin² x − 3sin x + 1 = 0

for:

0° ≤ x < 360°

Let:

u = sin x

Then:

2u² − 3u + 1 = 0

Factor:

(2u − 1)(u − 1) = 0

Therefore:

u = 1/2

or:

u = 1

Substitute back.

First Case

sin x = 1/2

x = 30°, 150°

Second Case

sin x = 1

x = 90°

Final Answer

x = 30°, 90°, 150°

General Solution

x = 30° + 360°n

or:

x = 150° + 360°n

or:

x = 90° + 360°n

where n ∈ ℤ.

Solving Quadratic Cosine Equations

Worked Example

Solve:

2cos² x + cos x − 1 = 0

for:

0° ≤ x < 360°

Let:

u = cos x

Then:

2u² + u − 1 = 0

Factor:

(2u − 1)(u + 1) = 0

Therefore:

u = 1/2

or:

u = −1

First Case

cos x = 1/2

x = 60°, 300°

Second Case

cos x = −1

x = 180°

Final Answer

x = 60°, 180°, 300°

Solving Equations Using Pythagorean Identities

Trigonometric identities can transform equations containing more than one trigonometric function.

Worked Example

Solve:

sin² x + cos x = 1

for:

0° ≤ x < 360°

Use:

sin² x = 1 − cos² x

Substitute:

1 − cos² x + cos x = 1

Subtract 1:

−cos² x + cos x = 0

Factor:

cos x(1 − cos x) = 0

Therefore:

cos x = 0

or:

cos x = 1

First Case

cos x = 0

x = 90°, 270°

Second Case

cos x = 1

x = 0°

Final Answer

x = 0°, 90°, 270°

Solving Equations Using Double-Angle Identities

Worked Example

Solve:

sin 2x = sin x

for:

0° ≤ x < 360°

Use:

sin 2x = 2sin x cos x

Substitute:

2sin x cos x = sin x

Move everything to one side:

2sin x cos x − sin x = 0

Factor:

sin x(2cos x − 1) = 0

Therefore:

sin x = 0

or:

cos x = 1/2

First Case

sin x = 0

x = 0°, 180°

Second Case

cos x = 1/2

x = 60°, 300°

Final Answer

x = 0°, 60°, 180°, 300°

Important Algebraic Lesson

Do not divide both sides by sin x.

Doing so would eliminate the solutions where sin x = 0.

Factoring preserves all possible solutions.

Solving Equations Using Triple-Angle Identities

Worked Example

Solve:

sin 3x = sin x

for:

0° ≤ x < 360°

Use:

sin 3x = 3sin x − 4sin³ x

Substitute:

3sin x − 4sin³ x = sin x

Rearrange:

2sin x − 4sin³ x = 0

Factor:

2sin x(1 − 2sin² x) = 0

Therefore:

sin x = 0

or:

sin² x = 1/2

First Case

sin x = 0

x = 0°, 180°

Second Case

sin² x = 1/2

Therefore:

sin x = ±√2/2

The solutions are:

45°, 135°, 225°, 315°

Final Answer

x = 0°, 45°, 135°, 180°, 225°, 315°

Solving Equations Involving Sine and Cosine

Some equations contain both sine and cosine.

Worked Example

Solve:

sin x = cos x

for:

0° ≤ x < 360°

Since cosine cannot be zero at a solution of this equation, divide both sides by cos x:

sin x/cos x = 1

Use the quotient identity:

tan x = 1

Tangent equals 1 at:

45° and 225°.

Therefore:

x = 45°, 225°

General Solution

x = 45° + 180°n

where n ∈ ℤ.

Solving Equations with Coefficients of Sine and Cosine

Worked Example

Solve:

√3sin x = cos x

for:

0° ≤ x < 360°

At a solution, cos x cannot be zero, so divide both sides by cos x:

√3(sin x/cos x) = 1

Use:

tan x = sin x/cos x

Then:

√3tan x = 1

Therefore:

tan x = 1/√3

Since:

tan 30° = 1/√3

the solutions are:

x = 30°, 210°

General Solution

x = 30° + 180°n

where n ∈ ℤ.

Solving Equations Using Sum-to-Product Identities

Sum-to-product identities are particularly useful when equations contain sums or differences of trigonometric functions.

Worked Example

Solve:

sin 5x + sin 3x = 0

for:

0° ≤ x < 360°

Use:

sin A + sin B = 2sin[(A + B)/2]cos[(A − B)/2]

Therefore:

sin 5x + sin 3x

= 2sin 4x cos x

The equation becomes:

2sin 4x cos x = 0

Apply the zero-product property:

sin 4x = 0

or:

cos x = 0

First Case

sin 4x = 0

4x = 180°n

x = 45°n

Within the interval:

x = 0°, 45°, 90°, 135°, 180°, 225°, 270°, 315°

Second Case

cos x = 0

x = 90°, 270°

These are already included.

Final Answer

x = 0°, 45°, 90°, 135°, 180°, 225°, 270°, 315°

Solving Equations over Larger Intervals

Some problems specify intervals covering several revolutions.

Worked Example

Solve:

sin x = 1/2

for:

0° ≤ x < 720°

The solutions in one revolution are:

30° and 150°.

Sine repeats every 360°.

Add 360°:

390° and 510°.

Therefore:

x = 30°, 150°, 390°, 510°

Worked Example

Solve:

cos x = 0

for:

−360° ≤ x < 360°

Cosine equals zero at:

90° + 180°n

Choose integer values of n that produce angles within the interval.

The solutions are:

x = −270°, −90°, 90°, 270°

Verifying Trigonometric Solutions

Every proposed solution should satisfy the original equation.

Verification is especially important when:

• Squaring both sides.

• Multiplying by an expression containing the variable.

• Using inverse trigonometric functions.

• Applying algebraic transformations that may introduce extraneous solutions.

• Working with complicated multiple-angle equations.

Worked Example

Verify the solutions of:

2sin x − 1 = 0

The proposed solutions are:

x = 30°, 150°

For x = 30°:

2sin 30° − 1

= 2(1/2) − 1

= 0

For x = 150°:

2sin 150° − 1

= 2(1/2) − 1

= 0

Both solutions satisfy the equation.

Extraneous Solutions

An extraneous solution is a value obtained during algebraic manipulation that does not satisfy the original equation.

This can happen when both sides are squared.

Worked Example

Solve:

sin x = cos² x

for:

0° ≤ x < 360°

Use:

cos² x = 1 − sin² x

Substitute:

sin x = 1 − sin² x

Rearrange:

sin² x + sin x − 1 = 0

Let:

u = sin x

Then:

u² + u − 1 = 0

Use the quadratic formula:

u = [−1 ± √(1 + 4)]/2

Therefore:

u = (−1 ± √5)/2

The two algebraic values are:

u₁ = (−1 + √5)/2 ≈ 0.6180

u₂ = (−1 − √5)/2 ≈ −1.6180

Since sine must lie between −1 and 1, the second value is impossible.

Therefore:

sin x = (√5 − 1)/2

The reference angle is approximately:

38.17°

Since sine is positive in Quadrants I and II:

x ≈ 38.17°, 141.83°

These solutions satisfy the original equation, allowing for rounding.

The rejected algebraic root is outside the range of sine; it is not a valid trigonometric value.

Interpreting Multiple Solutions Graphically

A trigonometric equation can be interpreted as the intersection of two graphs.

For example:

sin x = 1/2

can be represented by:

y = sin x

and:

y = 1/2

The solutions are the x-coordinates where the curves intersect.

For:

0° ≤ x < 360°

there are two intersections.

For:

0° ≤ x < 720°

there are four intersections.

Why Graphs Are Useful

Graphs help us:

• Estimate the number of solutions.

• Identify repeated solutions.

• Understand the effect of changing the interval.

• Recognise the periodic nature of trigonometric functions.

• Check whether calculated answers are reasonable.

However, graphical estimates should not replace exact algebraic solutions when exact answers are required.

Applications in Physics: Oscillatory Motion

Trigonometric equations are used to determine when an oscillating object reaches a particular position.

Suppose a mass moves according to:

x(t) = 0.20sin(2πt)

where x is displacement in metres and t is time in seconds.

Find the times during:

0 ≤ t < 2

when:

x = 0.10 m

Step 1: Form the Equation

0.20sin(2πt) = 0.10

Divide by 0.20:

sin(2πt) = 1/2

Step 2: Solve for the Angle

The general solutions are:

2πt = π/6 + 2πn

or:

2πt = 5π/6 + 2πn

Step 3: Solve for Time

Divide by 2π:

t = 1/12 + n

or:

t = 5/12 + n

Step 4: Apply the Interval

For:

0 ≤ t < 2

the solutions are:

t = 1/12, 5/12, 13/12, 17/12 seconds

Approximately:

t ≈ 0.083, 0.417, 1.083, 1.417 seconds

The object reaches the specified displacement twice during each oscillation.

Applications in Circular Motion

Suppose a point moves around a circle of radius 5 m.

Its vertical position is:

y = 5sin θ

Find all angles between 0° and 360° where:

y = 2.5 m

Substitute:

5sin θ = 2.5

Divide by 5:

sin θ = 1/2

Therefore:

θ = 30°, 150°

The point reaches the same vertical height at two different positions on the circle.

This demonstrates why multiple trigonometric solutions can represent different physical situations.

Applications in Navigation

Trigonometric equations can be used to determine possible directions.

Suppose a displacement has magnitude 100 km and an eastward component of 60 km.

The eastward component is:

x = 100cos θ

Therefore:

100cos θ = 60

Divide:

cos θ = 0.6

Find the reference angle:

θ ≈ 53.13°

For:

0° ≤ θ < 360°

cosine is positive in Quadrants I and IV.

Therefore:

θ ≈ 53.13°, 306.87°

These represent two possible directions with the same eastward component.

Additional information about the northward or southward component would be needed to choose between them.

A Strategy for Solving General Trigonometric Equations

When solving increasingly complex trigonometric equations, follow a systematic process.

1. Identify the interval. Determine whether the problem uses degrees or radians and whether endpoints are included.

2. Simplify the equation. Rearrange terms and apply algebraic rules.

3. Recognise the trigonometric structure. Look for sine, cosine, tangent, multiple angles, quadratic forms, or identities.

4. Choose an appropriate method. Use reference angles, factorisation, substitutions, or trigonometric identities.

5. Find the reference angle. Use exact special-angle values or inverse trigonometric functions.

6. Determine the correct quadrants. Use the signs of trigonometric functions.

7. Write the general solution. Include the appropriate period and integer n.

8. Apply the interval. Select every solution that satisfies the required boundaries.

9. Remove duplicates. Different branches may produce the same solution.

10. Check domain restrictions. Exclude angles where the original equation is undefined.

11. Verify solutions. Substitute answers into the original equation.

12. State the final answer clearly. Use exact values when possible and sensible rounding when necessary.

Common Mistakes

Finding only one solution

Sine and cosine equations often have two solutions per cycle.

Forgetting periodicity

General solutions require the appropriate integer multiple of the period.

Using 360° for tangent

Tangent repeats every 180°, so the most efficient general solution uses 180°n.

Ignoring the specified interval

A valid general solution may lie outside the required interval.

Including excluded endpoints

Pay attention to strict inequalities such as x < 360°.

Mixing degrees and radians

Use consistent angle units throughout a calculation.

Choosing the wrong quadrants

Use the sign of the trigonometric function to determine valid quadrants.

Forgetting negative solutions

General solutions include negative as well as positive angles.

Failing to expand the interval for multiple angles

For sin 3x, the argument covers three times the interval of x.

Dividing by a trigonometric expression that may equal zero

This can eliminate valid solutions.

Ignoring the range of sine and cosine

Sine and cosine cannot exceed 1 or fall below −1.

Forgetting both roots of a quadratic

Factorisation may produce multiple trigonometric equations.

Failing to verify answers

Algebraic manipulation may introduce invalid solutions.

Rounding too early

Keep exact values or sufficient decimal precision until the final answer.

Listing duplicate solutions

Different branches may produce the same angle.

Assuming every solution has the same physical meaning

Multiple solutions may represent different positions, times, or directions.

Key Terms

Trigonometric Equation: An equation containing one or more trigonometric functions of an unknown angle.

Trigonometric Identity: An equation involving trigonometric functions that is true wherever both sides are defined.

Solution: A value of the variable that makes an equation true.

General Solution: A formula describing all solutions of an equation, usually using an integer parameter.

Principal Value: A selected value returned by an inverse trigonometric function within its defined range.

Reference Angle: The acute angle between an angle's terminal side and the x-axis.

Quadrant: One of the four regions of the Cartesian coordinate plane.

Unit Circle: A circle of radius 1 centred at the origin.

Periodicity: The property of a function whose values repeat at regular intervals.

Period: The length of one complete cycle of a periodic function.

Coterminal Angles: Angles that share the same terminal side.

Specified Interval: The range of values within which solutions must be found.

Closed Interval: An interval that includes both endpoints.

Open Interval: An interval that excludes both endpoints.

General Sine Solution: A formula representing every angle with a specified sine value.

General Cosine Solution: A formula representing every angle with a specified cosine value.

General Tangent Solution: A formula representing every angle with a specified tangent value.

Inverse Sine: The function arcsin, used to determine a principal angle from a sine value.

Inverse Cosine: The function arccos, used to determine a principal angle from a cosine value.

Inverse Tangent: The function arctan, used to determine a principal angle from a tangent value.

Quadratic Trigonometric Equation: A trigonometric equation containing a squared trigonometric function.

Factorisation: Rewriting an expression as a product of factors.

Zero-Product Property: The principle that a product equals zero when at least one factor equals zero.

Double-Angle Identity: An identity expressing a trigonometric function of twice an angle.

Triple-Angle Identity: An identity expressing a trigonometric function of three times an angle.

Compound-Angle Identity: An identity involving the sum or difference of two angles.

Sum-to-Product Identity: An identity converting a sum or difference of trigonometric functions into a product.

Extraneous Solution: A proposed solution obtained during manipulation that does not satisfy the original equation.

Verification: Checking that a proposed solution satisfies the original equation.

Domain Restriction: A condition excluding values where an expression is undefined.

Range: The set of possible output values of a function.

Graphical Solution: A solution determined from the intersection of graphs.

Phase Shift: A horizontal displacement of a periodic function.

Angular Frequency: The rate of angular change, commonly represented by ω.

Oscillation: Repeated motion around an equilibrium position.

Sinusoidal Motion: Motion described using sine or cosine functions.

Mathematical Modelling: The use of mathematical relationships to represent real-world situations.

Check Your Understanding

1. Explain why trigonometric equations often have infinitely many solutions.

2. State the periods of sine, cosine, and tangent in degrees and radians.

3. Solve:

sin x = √3/2

for 0° ≤ x < 360°.

4. Solve:

cos x = −1/2

for 0° ≤ x < 360°.

5. Determine the general solution of:

tan x = 1

in degrees.

6. Determine the general solution of:

sin x = −√2/2

in radians.

7. Solve:

3sin x − 1 = 0

for 0° ≤ x < 360°. Give answers to two decimal places.

8. Explain why:

cos x = 1.5

has no real solutions.

9. Solve:

sin 2x = √3/2

for 0° ≤ x < 360°.

10. Solve:

cos 3x = 1/2

for 0° ≤ x < 360°.

11. Solve:

sin(x + 45°) = √2/2

for 0° ≤ x < 360°.

12. Solve:

2sin² x − 3sin x + 1 = 0

for 0° ≤ x < 360°.

13. Solve:

sin 2x = sin x

for 0° ≤ x < 360°.

14. Solve:

sin 5x − sin x = 0

for 0° ≤ x < 360°. Use an appropriate trigonometric identity.

15. A mass oscillates according to:

x(t) = 0.30cos(πt)

where x is measured in metres and t in seconds.

Determine every time in the interval 0 ≤ t < 4 when the displacement is 0.15 m. Explain why multiple solutions occur and verify your answers.

Key Takeaways

• Trigonometric equations are solved by finding every angle that satisfies the equation within the required domain or interval.

• Trigonometric functions are periodic, so equations often have infinitely many solutions.

• Sine and cosine repeat every 360° or 2π radians.

• Tangent repeats every 180° or π radians.

• Reference angles help determine solutions in different quadrants.

• The signs of sine, cosine, and tangent determine which quadrants contain valid solutions.

• General solutions use an integer parameter n to represent repeated cycles.

• Sine equations generally have two solution families per cycle, except at their maximum and minimum values.

• Cosine equations generally have two solution families per cycle, except at their maximum and minimum values.

• Tangent equations can be represented using one repeating solution family.

• Specified intervals determine which solutions must be included in the final answer.

• Open and closed interval endpoints must be handled carefully.

• Inverse trigonometric functions provide reference angles for values that are not standard special angles.

• Sine and cosine values must lie between −1 and 1.

• Equations involving multiple angles may have additional solutions because the functions complete more cycles.

• Compound-angle equations can be solved by isolating the complete angle expression.

• Quadratic trigonometric equations can often be solved using algebraic factorisation.

• Pythagorean identities allow equations involving sine and cosine to be rewritten in a single function.

• Double-angle and triple-angle identities are useful for solving more complex equations.

• Sum-to-product identities can transform trigonometric sums into factorable equations.

• Dividing by an expression containing the variable may eliminate valid solutions.

• Every proposed solution should be checked against the original equation.

• Graphs help interpret why multiple solutions occur.

• Trigonometric equations are useful in circular motion, oscillations, navigation, and other periodic systems.

• Mastering general trigonometric equations develops the algebraic reasoning, graphical understanding, and problem-solving skills needed for advanced mathematics, physics, and calculus.

2. Equations Involving Multiple Angles

3. Systems of Trigonometric Equations

Systems of Trigonometric Equations

A system of trigonometric equations consists of two or more equations involving trigonometric functions that must be satisfied simultaneously.

Unlike a single trigonometric equation, where we find values that satisfy one relationship, a system requires us to find values that satisfy every equation at the same time.

For example:

sin x + cos y = 1

sin x − cos y = 0

Both equations contain trigonometric expressions involving two unknown angles. To solve the system, we must find every ordered pair (x, y) that makes both equations true.

Systems of trigonometric equations can be solved using familiar algebraic techniques, including substitution and elimination. However, the periodic nature of trigonometric functions means that systems may have several solutions, infinitely many solutions, or no solutions.

These equations are useful in physics, engineering, navigation, wave analysis, and the study of periodic phenomena.

Understanding Simultaneous Trigonometric Equations

A simultaneous trigonometric system may involve one or more unknown angles.

For example:

System A

sin x = cos x

sin 2x = 1

This system contains one unknown angle, x, and requires a value that satisfies both equations.

System B

sin x + cos y = 1

sin x − cos y = 0

This system contains two unknown angles, x and y.

System C

sin x + sin y = 1

sin x − sin y = 0

This system contains two trigonometric expressions that can be isolated using algebraic elimination.

What Counts as a Solution?

For a system involving two variables, a solution is an ordered pair:

(x, y)

The first coordinate represents x, and the second represents y.

For example:

(30°, 60°)

and:

(60°, 30°)

are different ordered pairs.

Even when both pairs contain the same two angles, they may not both satisfy a particular system.

Important Principle

A valid solution must satisfy every equation in the system, not just one.

Reviewing the Algebraic Methods

The techniques used for linear simultaneous equations also apply to many trigonometric systems.

Substitution

Substitution involves isolating one variable or expression in an equation and replacing it in another equation.

For example:

sin x + cos y = 1

sin x = 2cos y

Substitute the second equation into the first:

2cos y + cos y = 1

3cos y = 1

cos y = 1/3

Once y is determined, substitute back to find x.

Elimination

Elimination involves adding or subtracting equations to remove one expression.

For example:

sin x + cos y = 1

sin x − cos y = 0

Add the equations:

2sin x = 1

sin x = 1/2

The system has been reduced to a simpler trigonometric equation.

Why These Methods Work

The expressions sin x and cos y can temporarily be treated as algebraic quantities.

For example, let:

a = sin x

b = cos y

Then:

a + b = 1

a − b = 0

This is an ordinary system of linear equations.

After solving for a and b, we return to the trigonometric equations to determine the angles.

Solving Systems by Elimination

Worked Example

Solve:

sin x + cos y = 1

sin x − cos y = 0

for:

0° ≤ x < 360°

0° ≤ y < 360°

Step 1: Add the Equations

sin x + cos y = 1

sin x − cos y = 0

Adding:

2sin x = 1

Therefore:

sin x = 1/2

Step 2: Solve for x

Sine is positive in Quadrants I and II.

Therefore:

x = 30°, 150°

Step 3: Find cos y

Substitute sin x = 1/2 into:

sin x − cos y = 0

Then:

1/2 − cos y = 0

Therefore:

cos y = 1/2

Cosine is positive in Quadrants I and IV.

Thus:

y = 60°, 300°

Step 4: Form All Ordered Pairs

Every permitted x-value can be paired with every permitted y-value because the reduced equations are independent.

The complete solution set is:

(30°, 60°)

(30°, 300°)

(150°, 60°)

(150°, 300°)

Step 5: Verify

For the pair (30°, 60°):

sin 30° + cos 60° = 1/2 + 1/2 = 1

sin 30° − cos 60° = 1/2 − 1/2 = 0

Both equations are satisfied.

The remaining pairs also satisfy both equations because they have the same required sine and cosine values.

Solving Systems by Substitution

Worked Example

Solve:

sin x + cos y = 1

sin x = 2cos y

for:

0° ≤ x < 360°

0° ≤ y < 360°

Step 1: Substitute

Replace sin x with 2cos y:

2cos y + cos y = 1

Therefore:

3cos y = 1

cos y = 1/3

Step 2: Solve for y

Let:

α = arccos(1/3)

α ≈ 70.53°

Cosine is positive in Quadrants I and IV.

Therefore:

y ≈ 70.53°, 289.47°

Step 3: Solve for x

From:

sin x = 2cos y

we obtain:

sin x = 2/3

Let:

β = arcsin(2/3)

β ≈ 41.81°

Sine is positive in Quadrants I and II.

Therefore:

x ≈ 41.81°, 138.19°

Step 4: Form the Solution Set

The four ordered pairs are approximately:

• (41.81°, 70.53°)

• (41.81°, 289.47°)

• (138.19°, 70.53°)

• (138.19°, 289.47°)

All four satisfy the system, subject to rounding.

Important Observation

The algebraic substitution produced one value for sin x and one value for cos y.

However, each trigonometric equation produced two possible angles.

This is why a system that appears to have one algebraic solution may have several trigonometric solutions.

Using Algebraic Variables to Simplify a System

A useful technique is to introduce temporary variables representing trigonometric expressions.

Worked Example

Solve:

2sin x + 3cos y = 2

sin x − cos y = 1/2

for:

0° ≤ x < 360°

0° ≤ y < 360°

Let:

a = sin x

b = cos y

The system becomes:

2a + 3b = 2

a − b = 1/2

Step 1: Isolate a

From the second equation:

a = b + 1/2

Step 2: Substitute

2(b + 1/2) + 3b = 2

Expand:

2b + 1 + 3b = 2

5b = 1

Therefore:

b = 1/5

Step 3: Find a

a = 1/5 + 1/2

a = 7/10

Therefore:

sin x = 0.7

cos y = 0.2

Step 4: Determine the Angles

For x:

x = arcsin(0.7)

or:

x = 180° − arcsin(0.7)

Approximately:

x ≈ 44.43°, 135.57°

For y:

y = arccos(0.2)

or:

y = 360° − arccos(0.2)

Approximately:

y ≈ 78.46°, 281.54°

Final Answer

The system has four solutions:

(44.43°, 78.46°)

(44.43°, 281.54°)

(135.57°, 78.46°)

(135.57°, 281.54°)

All values are approximate.

Recognising Impossible Systems

Not every algebraic solution corresponds to a valid trigonometric solution.

Recall:

−1 ≤ sin x ≤ 1

−1 ≤ cos x ≤ 1

Worked Example

Solve:

sin x + cos y = 3

sin x − cos y = 1

Add the equations:

2sin x = 4

sin x = 2

However, sine cannot equal 2 for any real angle.

Therefore:

The system has no real solutions.

General Principle

Always check that the values obtained for sine and cosine lie within their possible ranges.

Solving Systems with One Unknown Angle

Some simultaneous trigonometric equations involve only one variable.

In these cases, the solution must belong to the solution sets of both equations.

Worked Example

Solve:

sin x = 1/2

cos 2x = 1/2

for:

0° ≤ x < 360°

Step 1: Solve the First Equation

sin x = 1/2

Therefore:

x = 30°, 150°

Step 2: Check the Second Equation

For x = 30°:

cos 2x = cos 60° = 1/2

Valid.

For x = 150°:

cos 2x = cos 300° = 1/2

Valid.

Final Answer

x = 30°, 150°

Both values satisfy the complete system.

Alternative Method

Use:

cos 2x = 1 − 2sin² x

Since:

sin x = 1/2

then:

cos 2x = 1 − 2(1/2)²

= 1 − 1/2

= 1/2

Therefore, the second equation is automatically satisfied whenever the first equation is satisfied.

This reveals that the second equation imposes no additional restriction on the solutions of the first.

Solving Systems by Comparing Solution Sets

Worked Example

Solve:

sin x = cos x

sin 2x = 1

for:

0° ≤ x < 360°

Step 1: Solve the First Equation

sin x = cos x

Since cos x cannot equal zero at a solution, divide by cos x:

tan x = 1

Therefore:

x = 45°, 225°

Step 2: Solve the Second Equation

sin 2x = 1

2x = 90° + 360°n

Therefore:

x = 45° + 180°n

Within the interval:

x = 45°, 225°

Final Answer

x = 45°, 225°

Both equations produce the same solution set over the specified interval.

Systems with No Common Solutions

Worked Example

Solve:

sin x = 1/2

cos x = 1/2

for:

0° ≤ x < 360°

The first equation gives:

x = 30°, 150°

The second equation gives:

x = 60°, 300°

There are no common angles.

Therefore:

The system has no solution.

Algebraic Verification

If both equations were true, then:

sin² x + cos² x = (1/2)² + (1/2)²

= 1/4 + 1/4

= 1/2

But the Pythagorean identity requires:

sin² x + cos² x = 1

This contradiction confirms that no solution exists.

Systems Involving Sine and Cosine of the Same Angle

Systems containing sin x and cos x are especially interesting because the two expressions are not independent.

They must satisfy:

sin² x + cos² x = 1

Worked Example

Solve:

sin x + cos x = √2

sin x − cos x = 0

for:

0° ≤ x < 360°

Step 1: Add the Equations

2sin x = √2

Therefore:

sin x = √2/2

Step 2: Find cos x

From:

sin x − cos x = 0

we obtain:

cos x = √2/2

Step 3: Determine x

Both sine and cosine are positive in Quadrant I.

Therefore:

x = 45°

Verification

sin 45° + cos 45°

= √2/2 + √2/2

= √2

And:

sin 45° − cos 45° = 0

Both equations are satisfied.

Why Sine and Cosine of the Same Angle Cannot Be Chosen Independently

Suppose an algebraic system produces:

sin x = 0.8

cos x = 0.8

Both values lie individually within the interval from −1 to 1.

However:

sin² x + cos² x

= 0.64 + 0.64

= 1.28

This contradicts the Pythagorean identity.

Therefore, no real angle can satisfy both conditions.

Important Distinction

For a system involving sin x and cos y, the angles x and y are independent unless another equation relates them.

For a system involving sin x and cos x, the expressions are linked by the unit-circle identity.

This distinction determines whether separate angle solutions can be freely combined.

Solving Nonlinear Systems Using Trigonometric Identities

Some systems contain products or powers of trigonometric expressions.

Worked Example

Solve:

sin x + cos x = 1

sin x cos x = 0

for:

0° ≤ x < 360°

Step 1: Apply the Zero-Product Property

From:

sin x cos x = 0

we obtain:

sin x = 0

or:

cos x = 0

Case A: sin x = 0

The first equation becomes:

cos x = 1

Therefore:

x = 0°

Case B: cos x = 0

The first equation becomes:

sin x = 1

Therefore:

x = 90°

Final Answer

x = 0°, 90°

Verification

For x = 0°:

sin 0° + cos 0° = 0 + 1 = 1

sin 0° cos 0° = 0

For x = 90°:

sin 90° + cos 90° = 1 + 0 = 1

sin 90° cos 90° = 0

Both values satisfy the system.

Solving Systems Using Double-Angle Identities

Worked Example

Solve:

sin 2x = 0

cos x = −1

for:

0° ≤ x < 360°

Step 1: Solve the Second Equation

cos x = −1

Therefore:

x = 180°

Step 2: Verify the First Equation

sin 2(180°)

= sin 360°

= 0

Therefore:

x = 180°

is the only solution.

Alternative Method

Use:

sin 2x = 2sin x cos x

Then:

2sin x cos x = 0

Since cos x = −1, it is nonzero, so:

sin x = 0

Together with cos x = −1, this gives:

x = 180°

Systems with Two Unknown Angles and a Relationship Between Them

Some systems involve x and y through a direct relationship, such as:

x + y = 90°

This additional condition can significantly reduce the number of solutions.

Worked Example

Solve:

sin x = cos y

x + y = 90°

for:

0° ≤ x ≤ 90°

0° ≤ y ≤ 90°

Step 1: Use the Angle Relationship

From:

x + y = 90°

we obtain:

y = 90° − x

Step 2: Substitute

sin x = cos(90° − x)

Use the cofunction identity:

cos(90° − x) = sin x

Therefore:

sin x = sin x

This is an identity.

Interpretation

Every pair satisfying:

x + y = 90°

within the specified interval is a solution.

The complete solution set is:

(x, y) = (t, 90° − t)

where:

0° ≤ t ≤ 90°

Important Observation

This system has infinitely many solutions, even within the bounded interval.

The first equation does not provide any additional restriction beyond the second.

Solving Systems with a Sum of Angles

Worked Example

Solve:

sin x = cos y

x + y = 180°

for:

0° ≤ x < 180°

0° < y ≤ 180°

Step 1: Substitute

From:

x + y = 180°

we obtain:

y = 180° − x

Substitute:

sin x = cos(180° − x)

Use:

cos(180° − x) = −cos x

Therefore:

sin x = −cos x

Step 2: Solve

Since cos x cannot be zero at a solution:

tan x = −1

Within:

0° ≤ x < 180°

the solution is:

x = 135°

Step 3: Find y

y = 180° − 135°

y = 45°

Final Answer

(x, y) = (135°, 45°)

Verification

sin 135° = √2/2

cos 45° = √2/2

And:

135° + 45° = 180°

Both equations are satisfied.

Solving Systems with Quadratic Substitution

Some systems can be simplified by introducing:

a = sin x

b = cos y

and solving a nonlinear algebraic system.

Worked Example

Solve:

sin x + cos y = 1

sin x cos y = 3/16

for:

0° ≤ x < 360°

0° ≤ y < 360°

Step 1: Introduce Variables

Let:

a = sin x

b = cos y

Then:

a + b = 1

ab = 3/16

Step 2: Substitute

From:

a + b = 1

we obtain:

b = 1 − a

Substitute into:

ab = 3/16

Then:

a(1 − a) = 3/16

Expand:

a − a² = 3/16

Rearrange:

a² − a + 3/16 = 0

Multiply by 16:

16a² − 16a + 3 = 0

Factor:

(4a − 1)(4a − 3) = 0

Therefore:

a = 1/4

or:

a = 3/4

Step 3: Find b

If:

a = 1/4

then:

b = 3/4

If:

a = 3/4

then:

b = 1/4

Thus the possible trigonometric combinations are:

sin x = 1/4 and cos y = 3/4

or:

sin x = 3/4 and cos y = 1/4

Step 4: Determine the Angles

For sin x = 1/4:

x ≈ 14.48°, 165.52°

For cos y = 3/4:

y ≈ 41.41°, 318.59°

For sin x = 3/4:

x ≈ 48.59°, 131.41°

For cos y = 1/4:

y ≈ 75.52°, 284.48°

Final Answer

There are eight ordered pairs, formed by combining the two x-values with the two y-values within each valid algebraic case.

Each row produces four ordered pairs.

This example demonstrates how a nonlinear algebraic system can lead to several trigonometric solution sets.

General Solutions of Trigonometric Systems

When no interval is specified, solutions are often expressed using integer parameters.

Worked Example

Solve:

sin x = 1/2

cos y = √2/2

for all real x and y.

Step 1: General Solutions for x

sin x = 1/2

Therefore:

x = 30° + 360°m

or:

x = 150° + 360°m

Step 2: General Solutions for y

cos y = √2/2

Therefore:

y = 45° + 360°n

or:

y = 315° + 360°n

where:

m, n ∈ ℤ

Final Answer

The complete solution set consists of all combinations of the two x-families and two y-families.

The integers m and n are independent because x and y are independent variables.

Why Two Integer Parameters Are Needed

The value of x can repeat through one number of revolutions while y repeats through a different number.

Using the same integer for both variables would incorrectly exclude valid combinations.

Interpreting Multiple Solution Sets

A trigonometric system can have different types of solution sets.

Discrete and Continuous Solutions

A discrete solution set consists of separate individual solutions.

For example:

(30°, 60°)

(150°, 300°)

A continuous solution set contains an entire range of related solutions.

For example:

x + y = 90°

sin x = cos y

has the family:

(x, y) = (t, 90° − t)

for all permitted values of t.

Why This Matters

Two systems may both have infinitely many solutions but behave very differently.

One may have infinitely many isolated solutions due to periodic repetition.

Another may have infinitely many solutions forming a continuous curve.

Graphical Verification of Systems

Graphs provide a useful way to interpret and verify simultaneous trigonometric equations.

For equations involving one unknown, solutions are common intersection points.

For equations involving two unknowns, each equation defines a curve in the xy-plane. The simultaneous solutions occur where those curves intersect.

Example: One-Variable System

Consider:

sin x = cos x

sin 2x = 1

The first equation can be represented by the intersection of:

y = sin x

and:

y = cos x

The second equation can be represented by:

y = sin 2x

and:

y = 1

The valid solutions are the x-values that satisfy both relationships.

For:

0° ≤ x < 360°

the solutions are:

x = 45°, 225°

Graphical Interpretation

At x = 45°:

sin 45° = cos 45° = √2/2

sin 90° = 1

At x = 225°:

sin 225° = cos 225° = −√2/2

sin 450° = 1

Both points satisfy the system.

Graphing a Two-Variable System

Consider:

sin x + sin y = 1

sin x − sin y = 0

for:

0° ≤ x < 360°

0° ≤ y < 360°

Adding the equations gives:

2sin x = 1

Therefore:

sin x = 1/2

Subtracting gives:

2sin y = 1

Therefore:

sin y = 1/2

Thus:

x = 30°, 150°

y = 30°, 150°

The solutions are:

(30°, 30°)

(30°, 150°)

(150°, 30°)

(150°, 150°)

What Does the Graph Show?

The solution set contains four distinct ordered pairs.

Each point represents a combination of angles that satisfies both original equations.

The graph also demonstrates that finding two x-values and two y-values can produce four ordered pairs when the reduced conditions are independent.

Limitations of Graphical Verification

Graphs are useful for visual confirmation, but they may not reveal exact solutions.

For example, a graph may suggest:

x ≈ 41.8°

but an algebraic solution may be expressed exactly as:

x = arcsin(2/3)

Graphical methods should therefore support, rather than replace, algebraic reasoning.

Using Technology to Verify Solutions

Graphing calculators and mathematical software can help check trigonometric systems.

For a one-variable system, graph both equations and identify common solutions.

For a two-variable system, graph each equation as a relation between x and y.

Useful tools include:

• Desmos Graphing Calculator

• GeoGebra Graphing Calculator

Example Investigation

Enter:

y = sin x

y = cos x

Set the calculator to degree mode.

Observe the intersections over:

0° ≤ x < 360°

Then graph:

y = sin 2x

and:

y = 1

Compare the x-values satisfying both equations.

Important Reminder

A calculator must be set to the correct angle unit.

Entering angles in degrees while the calculator is in radian mode produces incorrect interpretations.

Practical Application: Resolving Forces

Simultaneous trigonometric equations can be used to determine the magnitudes of forces acting in different directions.

Suppose two forces act on an object.

Force A acts at 30° above the horizontal.

Force B acts at 60° above the horizontal.

Their horizontal and vertical components satisfy:

A cos 30° + B cos 60° = 100

A sin 30° + B sin 60° = 100

where A and B are measured in newtons.

Find the magnitudes of the two forces.

Step 1: Substitute Exact Values

Recall:

cos 30° = √3/2

sin 30° = 1/2

cos 60° = 1/2

sin 60° = √3/2

The equations become:

A√3/2 + B/2 = 100

A/2 + B√3/2 = 100

Multiply both equations by 2:

√3A + B = 200

A + √3B = 200

Step 2: Eliminate a Variable

From the first equation:

B = 200 − √3A

Substitute into the second:

A + √3(200 − √3A) = 200

Expand:

A + 200√3 − 3A = 200

Simplify:

−2A = 200 − 200√3

Therefore:

A = 100(√3 − 1)

Step 3: Find B

B = 200 − √3[100(√3 − 1)]

B = 200 − 300 + 100√3

Therefore:

B = 100(√3 − 1)

Final Answer

Both forces have the same magnitude:

A = B = 100(√3 − 1) N

Approximately:

A = B ≈ 73.21 N

Verification

Substituting A = B into the original equations gives the required horizontal and vertical totals of 100 N.

This application illustrates how trigonometric components can be combined with algebraic elimination.

Practical Application: Navigation

Suppose a boat travels along two separate paths.

The first displacement has magnitude a and direction 30° north of east.

The second displacement has magnitude b and direction 60° north of east.

The combined displacement is:

100 km east

and:

100 km north

The component equations are:

a cos 30° + b cos 60° = 100

a sin 30° + b sin 60° = 100

This is mathematically the same system as the previous force example.

Therefore:

a = b = 100(√3 − 1) km

Approximately:

a = b ≈ 73.21 km

Interpretation

The boat can achieve the specified displacement by travelling approximately 73.21 km in each of the two directions.

The same mathematical techniques apply to forces and navigation because both involve vector components.

Practical Application: Wave Interference

When two waves overlap, their displacements add.

Suppose two waves are represented by:

y₁ = A sin x

y₂ = A cos x

We want to find angles where their displacements are equal and their combined displacement is √2A.

The system is:

A sin x = A cos x

A sin x + A cos x = √2A

Assume A ≠ 0.

Divide both equations by A:

sin x = cos x

sin x + cos x = √2

Step 1: Use the First Equation

sin x = cos x

Step 2: Substitute into the Second

2sin x = √2

Therefore:

sin x = √2/2

Since sin x = cos x:

cos x = √2/2

Step 3: Determine x

Both functions are positive in Quadrant I.

Therefore:

x = 45° + 360°n

where n ∈ ℤ.

Interpretation

The two waves have equal positive displacements at these angles, producing the specified combined displacement.

Practical Application: Circular Motion

Suppose two objects move around a circular track.

Their vertical positions are:

y₁ = 5sin θ

y₂ = 5cos θ

Find the angles where the objects have equal vertical positions and both are above the horizontal axis.

Step 1: Set the Positions Equal

5sin θ = 5cos θ

Divide by 5:

sin θ = cos θ

Therefore:

tan θ = 1

The solutions within one revolution are:

θ = 45°, 225°

Step 2: Apply the Height Condition

At θ = 45°:

sin 45° = cos 45° = √2/2

Both positions are positive.

At θ = 225°:

sin 225° = cos 225° = −√2/2

Both positions are negative.

Therefore:

θ = 45°

is the only solution satisfying both conditions within one revolution.

Important Observation

A practical system may include inequalities or physical restrictions in addition to equations.

These restrictions help determine which mathematical solutions are meaningful.

A Systematic Strategy for Solving Trigonometric Systems

When solving simultaneous trigonometric equations, use the following process.

1. Identify the unknowns. Determine whether the system contains one angle or several.

2. Read the interval restrictions. Check the permitted values of each variable.

3. Examine the equation structure. Look for repeated expressions such as sin x or cos y.

4. Choose an algebraic method. Consider substitution, elimination, or introducing temporary variables.

5. Apply trigonometric identities when useful. Use Pythagorean, double-angle, or compound-angle identities to simplify the system.

6. Solve the resulting algebraic equations. Find the required trigonometric values.

7. Check possible ranges. Reject sine or cosine values outside −1 to 1.

8. Determine every angle. Use reference angles and quadrant rules.

9. Form ordered pairs carefully. Combine independent solutions only when permitted by the system.

10. Apply any additional relationships. Conditions such as x + y = 90° may restrict the pairs.

11. Remove duplicate solutions. Count each distinct solution once.

12. Verify every solution. Substitute into all original equations.

13. Use graphs when helpful. Confirm the number and approximate positions of solutions.

14. Interpret the results. Explain their meaning in practical contexts.

Common Mistakes

Solving only one equation

A valid solution must satisfy every equation in the system.

Finding only one trigonometric angle

A single sine or cosine value may correspond to two angles in one revolution.

Ignoring periodicity

General solutions may require integer multiples of 360° or 180°.

Combining dependent values incorrectly

sin x and cos x must satisfy the Pythagorean identity.

Failing to form all valid ordered pairs

Independent x- and y-solutions may produce several combinations.

Combining every possible pair without checking

Additional equations may restrict which combinations are valid.

Ignoring the range of sine and cosine

Values outside −1 to 1 cannot represent real sine or cosine values.

Using the wrong quadrant

Reference angles must be placed in quadrants with the correct trigonometric signs.

Mixing degrees and radians

Angle units must remain consistent.

Forgetting interval restrictions

Only solutions within the required intervals should be included.

Dividing by expressions that may equal zero

This can eliminate valid solutions.

Ignoring physical restrictions

Negative force magnitudes or impossible times may be inappropriate in practical contexts.

Assuming a graph proves an exact solution

Graphical methods usually provide approximate verification.

Failing to check both original equations

Every proposed solution must satisfy the complete system.

Rounding too early

Keep exact values until the final calculation when possible.

Key Terms

System of Trigonometric Equations: Two or more trigonometric equations that must be satisfied simultaneously.

Simultaneous Equations: Equations involving shared unknowns that are solved together.

Solution Set: The complete collection of values satisfying all equations.

Ordered Pair: A pair of values (x, y) in which the order identifies the value of each variable.

Substitution Method: An algebraic method in which an expression from one equation replaces an equivalent expression in another.

Elimination Method: An algebraic method that combines equations to remove an unknown or repeated expression.

Temporary Variable: A new variable introduced to simplify an algebraic or trigonometric expression.

Linear System: A system in which the unknown quantities occur to the first power and are not multiplied together.

Nonlinear System: A system containing products, powers, or other nonlinear relationships involving unknown quantities.

Trigonometric Identity: An equation involving trigonometric functions that is true wherever both sides are defined.

Pythagorean Identity: The relationship sin² x + cos² x = 1.

Double-Angle Identity: An identity involving trigonometric functions of twice an angle.

Compound-Angle Identity: An identity involving the sum or difference of angles.

Reference Angle: The acute angle between an angle's terminal side and the x-axis.

Quadrant: One of the four regions of the Cartesian coordinate plane.

Periodicity: The property of a function whose values repeat at regular intervals.

General Solution: A formula representing all solutions using integer parameters.

Independent Variables: Variables that can vary separately within the restrictions of a system.

Dependent Variables: Variables linked by an equation or other mathematical relationship.

Discrete Solution Set: A collection of separate individual solutions.

Continuous Solution Set: A family of solutions containing an uninterrupted range of values.

Graphical Verification: Checking solutions using intersections or relationships between graphs.

Intersection Point: A point where two curves meet.

Domain Restriction: A condition limiting the permitted input values of an expression.

Extraneous Solution: A candidate obtained during manipulation that does not satisfy the original system.

Zero-Product Property: The principle that a product equals zero when at least one factor equals zero.

Factorisation: Rewriting an expression as a product of factors.

Vector Component: The part of a vector acting along a specified direction.

Resultant Vector: A vector representing the combined effect of two or more vectors.

Wave Interference: The combination of two or more waves.

Circular Motion: Motion along a circular path.

Mathematical Modelling: The use of mathematical equations to represent real-world situations.

Verification: Checking that a proposed solution satisfies every original equation.

Check Your Understanding

1. Explain the difference between solving one trigonometric equation and solving a system of trigonometric equations.

2. Describe how substitution and elimination can be used to solve trigonometric systems.

3. Solve:

sin x + cos y = 1

sin x − cos y = 0

for 0° ≤ x < 360° and 0° ≤ y < 360°.

4. Solve:

sin x + sin y = 1

sin x − sin y = 0

for 0° ≤ x < 360° and 0° ≤ y < 360°.

5. Solve:

2sin x + cos y = 1

sin x − cos y = 0

for 0° ≤ x < 360° and 0° ≤ y < 360°.

6. Explain why the system:

sin x = 0.8

cos x = 0.8

has no real solution.

7. Solve:

sin x = 1/2

cos 2x = 1/2

for 0° ≤ x < 360°.

8. Solve:

sin x = cos x

sin 2x = 1

for 0° ≤ x < 360°.

9. Solve:

sin x + cos x = √2

sin x − cos x = 0

for 0° ≤ x < 360°.

10. Solve:

sin x + cos x = 1

sin x cos x = 0

for 0° ≤ x < 360°.

11. Solve:

sin x = cos y

x + y = 180°

for 0° ≤ x < 180° and 0° < y ≤ 180°.

12. Solve:

sin x + cos y = 1

sin x cos y = 3/16

for 0° ≤ x < 360° and 0° ≤ y < 360°.

13. Determine the general solutions of:

sin x = √3/2

cos y = −1/2

where x and y are measured in degrees.

14. Explain how graphs can be used to verify the solution set of a simultaneous trigonometric system. Describe one limitation of graphical verification.

15. Two forces A and B act at angles of 30° and 60° above the horizontal. Their resultant has a horizontal component of 80 N and a vertical component of 100 N.

Construct two simultaneous equations using horizontal and vertical components. Solve for A and B, verify your results, and explain the physical meaning of your answers.

Key Takeaways

• A system of trigonometric equations requires every equation to be satisfied simultaneously.

• Systems may involve one unknown angle or several unknown angles.

• Algebraic substitution and elimination are useful methods for solving trigonometric systems.

• Temporary variables can simplify systems containing repeated trigonometric expressions.

• A single sine or cosine value may correspond to multiple angles.

• Systems involving two independent angles may produce several ordered pairs.

• The order of values in an ordered pair matters.

• Independent trigonometric solutions can be combined when no additional relationship restricts them.

• Sine and cosine of the same angle are linked by sin² x + cos² x = 1.

• Algebraic solutions must be checked against the ranges of trigonometric functions.

• Some systems have one solution, several solutions, infinitely many solutions, or no solutions.

• General solutions use integer parameters to represent periodic repetition.

• Additional angle relationships can restrict the possible solution pairs.

• Trigonometric identities help simplify nonlinear systems.

• Double-angle identities are useful when a system contains both x and 2x.

• Graphical verification helps identify intersections and confirm solution sets.

• Graphs are generally less precise than exact algebraic solutions.

• Every proposed solution must be substituted into all original equations.

• Practical restrictions may eliminate mathematically valid but physically inappropriate solutions.

• Systems of trigonometric equations are useful in force analysis, navigation, circular motion, and wave interference.

• Mastering simultaneous trigonometric equations develops algebraic fluency, logical reasoning, graphical interpretation, and mathematical modelling skills essential for advanced mathematics, physics, and engineering.

4. Trigonometric Inequalities

Trigonometric Inequalities

Introduction to Trigonometric Inequalities

A trigonometric inequality is a mathematical statement that compares a trigonometric expression with a number or another expression using an inequality symbol.

Unlike trigonometric equations, which ask us to find specific angles that produce an exact value, trigonometric inequalities ask us to identify ranges of angles that satisfy a condition.

For example:

• sin θ > 0.5

• cos θ ≤ 0

• tan θ ≥ 1

• 2sin θ − 1 < 0

These inequalities are useful because many real-world situations involve conditions that must remain above or below a particular value rather than equal an exact value.

Examples include determining when a rotating object reaches a certain height, when daylight exceeds a particular duration, or when an alternating electrical signal is above a specified voltage.

Understanding Trigonometric Inequalities

The four main inequality symbols are:

The distinction between strict inequalities (> and <) and inclusive inequalities (≥ and ≤) is important.

For example:

• sin θ > 0.5 excludes angles where sin θ = 0.5.

• sin θ ≥ 0.5 includes angles where sin θ = 0.5.

These boundary angles determine whether the endpoints of solution intervals are included.

Why inequalities produce intervals

Consider:

sin θ > 0.5

Within one complete revolution, sin θ = 0.5 at:

θ = 30° and θ = 150°

Between these angles, the sine function has values greater than 0.5.

Therefore:

30° < θ < 150°

In interval notation:

(30°, 150°)

The endpoints are excluded because the inequality is strict.

Solving Sine Inequalities

Sine inequalities can be solved using the unit circle or the graph of y = sin θ.

The sine function has the following properties:

• Range: −1 ≤ sin θ ≤ 1

• Period: 360° or 2π radians

• Positive in Quadrants I and II

• Negative in Quadrants III and IV

Worked Example: Solving sin θ ≥ 0.5

Solve for 0° ≤ θ < 360°.

Step 1: Find the boundary angles.

sin θ = 0.5

θ = 30° and 150°

Step 2: Determine where the sine function satisfies the inequality.

Sine is greater than or equal to 0.5 between these angles.

Step 3: Write the solution.

30° ≤ θ ≤ 150°

Interval notation:

[30°, 150°]

Square brackets indicate that both endpoints are included.

Worked Example: Solving sin θ < −0.5

Solve for 0° ≤ θ < 360°.

The boundary angles are:

θ = 210° and 330°

The sine function is less than −0.5 between these angles.

Therefore:

210° < θ < 330°

Interval notation:

(210°, 330°)

Solving Cosine Inequalities

The cosine function has these properties:

• Range: −1 ≤ cos θ ≤ 1

• Period: 360° or 2π radians

• Positive in Quadrants I and IV

• Negative in Quadrants II and III

Cosine inequalities may produce two separate solution intervals within a complete revolution.

Worked Example: Solving cos θ > 0.5

Solve for 0° ≤ θ < 360°.

Step 1: Find the boundary angles.

cos θ = 0.5

θ = 60° and 300°

Step 2: Identify where cosine exceeds 0.5.

This occurs near the beginning and end of the revolution.

Step 3: Express the solution.

0° ≤ θ < 60° or 300° < θ < 360°

Interval notation:

[0°, 60°) ∪ (300°, 360°)

The symbol ∪ means union, indicating that either interval satisfies the inequality.

Notice that 0° is included because cos 0° = 1, while 360° is excluded because the original domain excludes it.

Solving Tangent Inequalities

The tangent function behaves differently from sine and cosine.

Important properties include:

• Period: 180° or π radians

• Range: all real numbers

• Undefined at θ = 90° + 180°k, where k is an integer

• Positive in Quadrants I and III

• Negative in Quadrants II and IV

Because tangent has vertical asymptotes, particular care is needed when determining solution intervals.

Worked Example: Solving tan θ ≥ 1

Solve for 0° ≤ θ < 360°.

Step 1: Find the boundary angles.

tan θ = 1

θ = 45° and 225°

Step 2: Identify where tangent is at least 1.

Tangent increases from 1 toward positive infinity in two intervals.

Step 3: Write the solution.

45° ≤ θ < 90° or 225° ≤ θ < 270°

Interval notation:

[45°, 90°) ∪ [225°, 270°)

The angles 90° and 270° must be excluded because tangent is undefined there.

Interpreting Solution Intervals

A solution interval describes a continuous range of values that satisfy an inequality.

There are four common forms of interval notation.

Multiple solution intervals

Some trigonometric inequalities have solutions in more than one interval.

For example:

cos θ < 0

For 0° ≤ θ < 360°:

90° < θ < 270°

Interval notation:

(90°, 270°)

However:

sin θ < 0.5

For the same domain:

0° ≤ θ < 30° or 150° < θ < 360°

Interval notation:

[0°, 30°) ∪ (150°, 360°)

Understanding these intervals allows us to describe precisely which angles satisfy a condition.

Representing Solutions Graphically

Graphs provide a visual method for solving trigonometric inequalities.

The general process is:

1. Graph the trigonometric function.

2. Draw a horizontal line representing the boundary value.

3. Identify the intersection points.

4. Determine where the graph lies above or below the boundary.

5. Express the corresponding x-values as intervals.

Worked Example: Graphical Solution

Solve:

2sin θ − 1 > 0

For 0° ≤ θ < 360°.

Step 1: Rearrange the inequality.

2sin θ > 1

sin θ > 0.5

Step 2: Identify the boundary angles.

θ = 30° and 150°

Step 3: Interpret the graph.

The sine curve lies above y = 0.5 between these angles.

Step 4: Write the solution.

30° < θ < 150°

Interval notation:

(30°, 150°)

The graphical and algebraic methods produce the same answer.

Solving Inequalities in Radians

Trigonometric inequalities can also be expressed in radians.

Important angle conversions include:

Worked Example: Solving sin x ≥ √2/2

For 0 ≤ x < 2π:

The boundary angles are:

x = π/4 and x = 3π/4

Sine is greater than or equal to √2/2 between these angles.

Therefore:

π/4 ≤ x ≤ 3π/4

Interval notation:

[π/4, 3π/4]

General Solutions and Periodicity

Trigonometric functions repeat their values at regular intervals.

This means that inequalities can have infinitely many solutions unless the domain is restricted.

For sine and cosine, the period is 360° or 2π.

For tangent, the period is 180° or π.

Worked Example: General Solution

Solve:

sin θ > 0.5

For all real values of θ.

Within one revolution:

30° < θ < 150°

Since sine repeats every 360°, the general solution is:

30° + 360°k < θ < 150° + 360°k

where k is any integer.

In radians:

π/6 + 2πk < θ < 5π/6 + 2πk

where k ∈ ℤ.

This notation represents all possible solution intervals.

Compound Trigonometric Inequalities

A compound inequality contains two conditions that must be satisfied simultaneously.

For example:

0 ≤ sin θ ≤ 0.5

This means:

sin θ ≥ 0

and

sin θ ≤ 0.5

Both conditions must be true.

Worked Example

Solve for 0° ≤ θ < 360°:

0 ≤ sin θ ≤ 0.5

Step 1: Identify where sin θ ≥ 0.

0° ≤ θ ≤ 180°

Step 2: Identify where sin θ ≤ 0.5.

0° ≤ θ ≤ 30° or 150° ≤ θ < 360°

Step 3: Find the intersection of the solution sets.

0° ≤ θ ≤ 30° or 150° ≤ θ ≤ 180°

Interval notation:

[0°, 30°] ∪ [150°, 180°]

The solution must satisfy both conditions, so we use the intersection of the two sets of possible values.

Real-World Applications

Trigonometric inequalities are especially useful when quantities change periodically.

Application: Height of a Ferris Wheel

A Ferris wheel has a radius of 10 m, and its centre is 12 m above the ground.

The height of a passenger is modeled by:

h(θ) = 12 + 10sin θ

where θ is measured in degrees.

Determine the angles for which the passenger is more than 17 m above the ground during one revolution.

Step 1: Write the inequality.

12 + 10sin θ > 17

Step 2: Rearrange.

10sin θ > 5

sin θ > 0.5

Step 3: Solve.

30° < θ < 150°

Interpretation:

The passenger is above 17 m between 30° and 150° of the wheel's rotation.

This represents one-third of a complete revolution.

Application: Alternating Electrical Voltage

An alternating voltage can be modeled by:

V(t) = 120sin(2πt)

where t is measured in seconds.

Determine when the voltage exceeds 60 V during the first second.

Step 1: Write the inequality.

120sin(2πt) > 60

Step 2: Simplify.

sin(2πt) > 0.5

Step 3: Solve the corresponding angle inequality.

π/6 < 2πt < 5π/6

Step 4: Divide by 2π.

1/12 < t < 5/12

Answer:

The voltage exceeds 60 V for:

0.0833 s < t < 0.4167 s

This method is useful when analyzing electrical signals and determining how long a voltage remains above a threshold.

Common Mistakes

Confusing equations with inequalities

An equation identifies boundary angles, while an inequality identifies intervals of angles.

For example, sin θ = 0.5 has two solutions in one complete revolution, but sin θ > 0.5 has an entire interval of solutions.

Using incorrect interval brackets

Strict inequalities exclude their boundary values.

• or < requires parentheses.

• ≥ or ≤ may require square brackets when the boundary is defined and included.

Forgetting the domain

A solution for 0° ≤ θ < 360° is different from a general solution covering all real angles.

Ignoring tangent asymptotes

Tangent is undefined at 90°, 270°, and equivalent angles. These values cannot be included in a solution.

Selecting the wrong side of a graph

For y > a, choose regions above the horizontal line y = a.

For y < a, choose regions below it.

Confusing union and intersection

• Union (∪) combines alternative solution intervals.

• Intersection (∩) identifies values satisfying multiple conditions simultaneously.

Forgetting to check the calculator mode

A calculator set to radians may give incorrect results when working with degree-based problems, and vice versa.

Key Terms

Key Takeaways

• Trigonometric inequalities describe ranges of angles, rather than only individual solutions.

• Solving an inequality usually begins by finding the angles where the corresponding equation is satisfied.

• The unit circle and trigonometric graphs help determine which intervals satisfy an inequality.

• Strict inequalities exclude boundary angles, while inclusive inequalities include valid boundary angles.

• Interval notation provides a concise and precise way to communicate solutions.

• A union combines separate solution intervals, while an intersection identifies values satisfying multiple conditions.

• Sine and cosine repeat every 360° or 2π radians, while tangent repeats every 180° or π radians.

• Tangent inequalities require special attention because tangent is undefined at its vertical asymptotes.

• General solutions use an integer parameter to describe repeated solution intervals.

• Trigonometric inequalities are useful for analyzing periodic phenomena such as rotating machinery, alternating current, oscillations, and seasonal cycles.

Check Your Understanding

Recall and Understanding

1. What is the difference between a trigonometric equation and a trigonometric inequality?

2. Explain why sin θ > 0.5 produces an interval rather than two individual angles.

3. What is the difference between (30°, 150°) and [30°, 150°]?

4. Why must 90° and 270° be excluded from solutions involving tan θ?

5. What does the union symbol ∪ mean in interval notation?

Application

6. Solve sin θ ≥ 0 for 0° ≤ θ < 360°.

7. Solve cos θ < 0 for 0° ≤ θ < 360°.

8. Solve tan θ > 1 for 0° ≤ θ < 180°.

9. Solve 2cos θ − 1 ≤ 0 for 0° ≤ θ < 360°.

10. Solve sin x < √3/2 for 0 ≤ x < 2π.

Analysis and Challenge

11. Explain why cos θ > 0.5 produces two separate intervals when the domain is 0° ≤ θ < 360°.

12. Solve the compound inequality 0 ≤ cos θ ≤ 0.5 for one complete revolution.

13. Find the general solution of sin θ ≤ −0.5, expressing your answer using an integer k.

14. A Ferris wheel passenger's height is modeled by h(θ) = 15 + 12sin θ. Determine the intervals when the passenger is at least 21 m above the ground.

15. An alternating electrical signal is modeled by V(t) = 80sin(4πt). Determine when the voltage is greater than 40 V during 0 ≤ t ≤ 1 second.

Final Summary

Trigonometric inequalities extend the methods used to solve trigonometric equations by identifying complete ranges of acceptable values. Rather than stopping at the angles where a function equals a particular value, we use graphs, unit-circle relationships, and interval notation to determine where the function satisfies the required condition.

Understanding how to interpret solution intervals, represent solutions graphically, and account for periodicity allows us to solve both mathematical inequalities and practical problems involving repeating patterns.

These techniques form an important foundation for further study of trigonometric functions, mathematical modeling, calculus, and periodic systems.

5. Mixed Trigonometric Problems