4. Solving Trigonometric Equations

Solving Trigonometric Equations

Trigonometric equations are equations that contain one or more trigonometric functions, such as sine, cosine, or tangent, with an unknown angle.

For example:

sin θ = 1/2

Unlike many algebraic equations, trigonometric equations often have multiple solutions because trigonometric functions are periodic. Their values repeat at regular intervals.

For example:

sin 30° = 1/2

sin 150° = 1/2

sin 390° = 1/2

All three angles satisfy the same equation.

Solving trigonometric equations therefore requires more than finding a single angle. We must understand the unit circle, use inverse trigonometric functions, identify solutions in different quadrants, and account for periodicity.

These skills are essential in mathematics, physics, engineering, navigation, and the study of waves and oscillations.

Understanding Trigonometric Equations

A trigonometric equation contains a trigonometric function of an unknown angle.

Examples include:

• sin x = 1/2

• cos x = −√2/2

• tan x = √3

• 2sin x − 1 = 0

• sin² x + sin x = 0

• cos 2x = 1/2

The goal is to determine the values of x that make the equation true.

A trigonometric equation may have:

• No solutions.

• One solution within a specified interval.

• Several solutions within a specified interval.

• Infinitely many solutions when no interval is specified.

Equations Compared with Identities

A trigonometric equation is true only for particular values of x.

For example:

sin x = 1/2

A trigonometric identity is true for every value where both sides are defined.

For example:

sin² x + cos² x = 1

Identities are often used to simplify equations before solving them.

Why Trigonometric Equations Have Multiple Solutions

Trigonometric functions are periodic.

The sine and cosine functions repeat every 360°, or 2π radians.

The tangent function repeats every 180°, or π radians.

Therefore:

sin(x + 360°) = sin x

cos(x + 360°) = cos x

tan(x + 180°) = tan x

For example:

sin 30° = sin 390° = sin 750° = 1/2

These angles differ by complete revolutions.

General Solutions

When no interval is specified, solutions are often written using an integer n.

For example:

x = 30° + 360°n

where n is any integer.

This represents infinitely many coterminal angles.

However, sine also equals 1/2 at 150°.

Therefore, the complete general solution is:

x = 30° + 360°n

or:

x = 150° + 360°n

where n is any integer.

Solving Basic Sine Equations

Consider:

sin x = 1/2

for:

0° ≤ x < 360°

Find the Reference Angle

Use inverse sine:

x = sin⁻¹(1/2)

The principal value is:

30°

Determine the Quadrants

Sine is positive in Quadrants I and II.

Therefore:

First solution:

x = 30°

Second solution:

x = 180° − 30°

x = 150°

Final Answer

x = 30°, 150°

Verifying the Solutions

sin 30° = 1/2

sin 150° = 1/2

Both values satisfy the original equation.

Solving Sine Equations with Negative Values

Consider:

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 negative in Quadrants III and IV.

Therefore:

Quadrant III:

x = 180° + 60°

x = 240°

Quadrant IV:

x = 360° − 60°

x = 300°

Final Answer

x = 240°, 300°

Verification

sin 240° = −√3/2

sin 300° = −√3/2

Both solutions are correct.

Solving Basic Cosine Equations

Consider:

cos x = 1/2

for:

0° ≤ x < 360°

Find the Reference Angle

cos⁻¹(1/2) = 60°

Determine the Quadrants

Cosine is positive in Quadrants I and IV.

First solution:

x = 60°

Second solution:

x = 360° − 60°

x = 300°

Final Answer

x = 60°, 300°

Worked Example: Negative Cosine

Solve:

cos x = −√2/2

for:

0° ≤ x < 360°

The reference angle is 45°.

Cosine is negative in Quadrants II and III.

Therefore:

x = 180° − 45° = 135°

x = 180° + 45° = 225°

x = 135°, 225°

Solving Basic Tangent Equations

Consider:

tan x = 1

for:

0° ≤ x < 360°

Find the Reference Angle

tan⁻¹(1) = 45°

Determine the Quadrants

Tangent is positive in Quadrants I and III.

Therefore:

x = 45°

x = 180° + 45° = 225°

Final Answer

x = 45°, 225°

Tangent Periodicity

Tangent repeats every 180°.

Therefore, the general solution is:

x = 45° + 180°n

where n is any integer.

Worked Example: Negative Tangent

Solve:

tan x = −√3

for:

0° ≤ x < 360°

The reference angle is 60°.

Tangent is negative in Quadrants II and IV.

Therefore:

x = 180° − 60° = 120°

x = 360° − 60° = 300°

x = 120°, 300°

Using the Unit Circle to Find Solutions

The unit circle provides a useful way to determine which quadrants contain solutions.

For an angle x:

• cos x is the horizontal coordinate.

• sin x is the vertical coordinate.

• tan x = sin x/cos x, where cosine is nonzero.

The signs of the functions depend on the quadrant.

A useful memory rule is ASTC:

• Quadrant I: All positive.

• Quadrant II: Sine positive.

• Quadrant III: Tangent positive.

• Quadrant IV: Cosine positive.

Finding Solutions from Reference Angles

If the reference angle is α:

These formulas help locate solutions within one complete revolution.

Solving Equations in Radians

Trigonometric equations can also be solved using radians.

A complete revolution is:

2π radians

A half revolution is:

π radians

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:

First solution:

x = π/4

Second solution:

x = π − π/4

x = 3π/4

Final Answer

x = π/4, 3π/4

Worked Example

Solve:

cos x = −1/2

for:

0 ≤ x < 2π

The reference angle is:

π/3

Cosine is negative in Quadrants II and III.

Therefore:

x = 2π/3

x = 4π/3

x = 2π/3, 4π/3

General Solutions for Sine, Cosine, and Tangent

When no interval is specified, we must include all possible solutions.

Let α be the principal angle returned by the appropriate inverse function.

Sine

If:

sin x = a

then:

x = α + 2πn

or:

x = π − α + 2πn

where:

α = arcsin(a)

and n is any integer.

This requires −1 ≤ a ≤ 1.

Cosine

If:

cos x = a

then:

x = ±α + 2πn

where:

α = arccos(a)

and n is any integer.

This requires −1 ≤ a ≤ 1.

Tangent

If:

tan x = a

then:

x = α + πn

where:

α = arctan(a)

and n is any integer.

General Solution Table

For sine and cosine, α is defined using their respective inverse functions.

These formulas can also be written in degrees by replacing 2π with 360° and π with 180°.

Solving Equations with Coefficients

Many trigonometric equations require algebraic rearrangement before finding the angle.

Worked Example

Solve:

2sin x − 1 = 0

for:

0° ≤ x < 360°

Isolate the Trigonometric Function

2sin x = 1

sin x = 1/2

Find the Solutions

The reference angle is 30°.

Sine is positive in Quadrants I and II.

Therefore:

x = 30°, 150°

Worked Example

Solve:

3cos x + 1 = 0

for:

0° ≤ x < 360°

Rearrange:

3cos x = −1

cos x = −1/3

Find the principal angle:

x = cos⁻¹(−1/3)

x ≈ 109.47°

Cosine is negative in Quadrants II and III.

The second solution is:

360° − 109.47° = 250.53°

Therefore:

x ≈ 109.47°, 250.53°

Equations with No Real Solutions

Not every trigonometric equation has a solution.

For example:

sin x = 2

Sine values always lie between −1 and 1.

Therefore, no real angle can satisfy this equation.

Similarly:

cos x = −1.5

has no real solutions.

Worked Example

Solve:

4sin x − 6 = 0

Rearrange:

4sin x = 6

sin x = 1.5

Since:

−1 ≤ sin x ≤ 1

the equation has:

No real solutions

Important Principle

Always check whether the required sine or cosine value is within the permitted range before calculating angles.

Tangent has no corresponding range restriction because tangent can take any real value.

Solving Equations Involving Multiple Angles

Some equations contain expressions such as:

sin 2x

cos 3x

tan(x − 30°)

These require careful consideration of the angle interval.

Worked Example

Solve:

sin 2x = 1/2

for:

0° ≤ x < 360°

Adjust the Interval

Since:

0° ≤ x < 360°

multiply the interval by 2:

0° ≤ 2x < 720°

Find All Solutions for 2x

Sine equals 1/2 at:

30°, 150°, 390°, 510°

within this interval.

Therefore:

2x = 30°, 150°, 390°, 510°

Divide each by 2:

x = 15°, 75°, 195°, 255°

Why Are There Four Solutions?

The function sin 2x completes two full cycles between 0° and 360°.

Therefore, there are more solutions than for sin x = 1/2 over the same interval.

Solving Cosine Equations with Multiple Angles

Worked Example

Solve:

cos 3x = 0

for:

0° ≤ x < 360°

Adjust the Interval

0° ≤ 3x < 1080°

Cosine equals zero at:

90°, 270°, 450°, 630°, 810°, 990°

Therefore:

3x = 90°, 270°, 450°, 630°, 810°, 990°

Divide by 3:

x = 30°, 90°, 150°, 210°, 270°, 330°

There are six solutions.

Solving Equations with Horizontal Translations

Some equations contain a phase shift.

Worked Example

Solve:

sin(x − 30°) = 1/2

for:

0° ≤ x < 360°

Let:

u = x − 30°

Then:

sin u = 1/2

The general solutions are:

u = 30° + 360°n

or:

u = 150° + 360°n

Substitute back:

x − 30° = 30° + 360°n

x = 60° + 360°n

or:

x − 30° = 150° + 360°n

x = 180° + 360°n

Within the interval:

x = 60°, 180°

Worked Example

Solve:

cos(x + 45°) = √2/2

for:

0° ≤ x < 360°

Let:

u = x + 45°

The general solutions are:

u = 45° + 360°n

or:

u = −45° + 360°n

Therefore:

x = 0° + 360°n

or:

x = −90° + 360°n

Within the interval:

x = 0°, 270°

Solving Equations Using Pythagorean Identities

Trigonometric identities can simplify equations containing several trigonometric functions.

Recall:

sin² x + cos² x = 1

Other useful forms include:

1 + tan² x = sec² x

1 + cot² x = csc² x

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 by Factoring

Factoring is a useful technique when an equation contains products or powers of trigonometric functions.

Worked Example

Solve:

sin² x − sin x = 0

for:

0° ≤ x < 360°

Factor:

sin x(sin x − 1) = 0

Therefore:

sin x = 0

or:

sin x − 1 = 0

First Case

sin x = 0

x = 0°, 180°

Second Case

sin x = 1

x = 90°

Final Answer

x = 0°, 90°, 180°

Important Warning

Do not divide both sides by sin x.

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

Factoring preserves all possible solutions.

Solving Quadratic Trigonometric Equations

Some trigonometric equations resemble quadratic equations.

For example:

2sin² x − 3sin x + 1 = 0

These can be solved using substitution or factoring.

Worked Example

Solve:

2sin² x − 3sin x + 1 = 0

for:

0° ≤ x < 360°

Treat Sine as a Variable

Let:

u = sin x

Then:

2u² − 3u + 1 = 0

Factor:

(2u − 1)(u − 1) = 0

Therefore:

u = 1/2

or:

u = 1

Substitute Back

sin x = 1/2

or:

sin x = 1

Find the Solutions

For sin x = 1/2:

x = 30°, 150°

For sin x = 1:

x = 90°

Final Answer

x = 30°, 90°, 150°

Solving Quadratic Equations Involving Cosine

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

Substitute back:

cos x = 1/2

or:

cos x = −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 Double-Angle Identities

Double-angle identities allow expressions involving 2x to be rewritten using functions of x.

Recall:

sin 2x = 2sin x cos x

cos 2x = 1 − 2sin² x

cos 2x = 2cos² x − 1

Worked Example

Solve:

sin 2x = sin x

for:

0° ≤ x < 360°

Use:

sin 2x = 2sin x cos x

Therefore:

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°

Solving Equations Using Cosine Double-Angle Identities

Worked Example

Solve:

cos 2x = cos x

for:

0° ≤ x < 360°

Use:

cos 2x = 2cos² x − 1

Substitute:

2cos² x − 1 = cos x

Rearrange:

2cos² x − cos x − 1 = 0

Factor:

(2cos x + 1)(cos x − 1) = 0

Therefore:

cos x = −1/2

or:

cos x = 1

First Case

cos x = −1/2

x = 120°, 240°

Second Case

cos x = 1

x = 0°

Final Answer

x = 0°, 120°, 240°

Solving Equations with Multiple Trigonometric Functions

Some equations contain sine and cosine together.

These may require factoring, division, or identities.

Worked Example

Solve:

sin x = cos x

for:

0° ≤ x < 360°

Where cos x ≠ 0, divide both sides by cos x:

sin x/cos x = 1

Therefore:

tan x = 1

The reference angle is 45°.

Tangent is positive in Quadrants I and III.

Therefore:

x = 45°, 225°

At angles where cos x = 0, sin x is ±1, so none of those excluded angles satisfy the original equation.

Worked Example

Solve:

√3 sin x = cos x

for:

0° ≤ x < 360°

Where cos x ≠ 0, divide by cos x:

√3 tan x = 1

Therefore:

tan x = 1/√3

tan x = √3/3

The reference angle is 30°.

Tangent is positive in Quadrants I and III.

Therefore:

x = 30°, 210°

Angles where cos x = 0 do not satisfy the original equation.

Verifying Trigonometric Solutions

Finding possible solutions is not always enough.

We should substitute them into the original equation to verify that they are valid.

Worked Example

Verify the solutions of:

2sin x − 1 = 0

We found:

x = 30°, 150°

Check x = 30°

2sin 30° − 1

= 2(1/2) − 1

= 1 − 1

= 0

The solution is valid.

Check x = 150°

2sin 150° − 1

= 2(1/2) − 1

= 0

The solution is valid.

Why Verification Matters

Verification can detect:

• Algebraic mistakes.

• Incorrect quadrant choices.

• Solutions outside the required interval.

• Invalid solutions introduced by squaring.

• Errors caused by dividing by zero.

Extraneous Solutions

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

These may appear when squaring both sides of an equation.

Worked Example

Solve:

sin x = cos² x

for:

0° ≤ x < 360°

Use:

cos² x = 1 − sin² x

Therefore:

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 ± √5)/2

The two values are:

u = (−1 + √5)/2 ≈ 0.6180

u = (−1 − √5)/2 ≈ −1.6180

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

Therefore:

sin x ≈ 0.6180

The reference angle is approximately:

38.17°

Sine is positive in Quadrants I and II.

Therefore:

x ≈ 38.17°, 141.83°

The rejected algebraic value is outside the range of sine and cannot produce a real solution.

Solving Equations Over Different Intervals

The interval determines which solutions must be included.

Consider:

sin x = 1/2

Interval: 0° ≤ x < 360°

Solutions:

30°, 150°

Interval: 0° ≤ x < 720°

Solutions:

30°, 150°, 390°, 510°

Interval: −180° ≤ x < 180°

Solutions:

−30°?

No. Since sin(−30°) = −1/2, it is not a solution.

The solutions in this interval are:

−210°?

This is outside the interval.

Instead, the correct solutions are:

30°, 150°

Interval: −360° ≤ x < 0°

Subtract 360° from the positive solutions:

30° − 360° = −330°

150° − 360° = −210°

Therefore:

x = −330°, −210°

Important Principle

Always check interval boundaries carefully.

If the interval uses:

0° ≤ x < 360°

then 0° is included, but 360° is excluded.

Solving Trigonometric Equations Graphically

Trigonometric equations can be interpreted as intersections of 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 two graphs intersect.

Graphical Method

To solve an equation graphically:

1. Graph the left-hand side.

2. Graph the right-hand side.

3. Identify the intersection points.

4. Read the corresponding x-values.

5. Check that the solutions lie within the required interval.

Advantages

Graphical methods are useful for:

• Estimating solutions.

• Checking algebraic answers.

• Understanding periodicity.

• Identifying the number of solutions.

• Solving equations that cannot easily be rearranged.

Limitations

Graphical solutions may be approximate.

For exact answers, algebraic methods and trigonometric identities are usually preferred.

Applications in Physics: Oscillatory Motion

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

Suppose a vibrating object has displacement:

x(t) = 5sin(πt/2)

where:

• x is displacement in centimetres.

• t is time in seconds.

Find when the displacement is 2.5 cm during the first 4 seconds.

Form the Equation

5sin(πt/2) = 2.5

Divide by 5:

sin(πt/2) = 1/2

Solve for the Angle

Let:

u = πt/2

Since 0 ≤ t ≤ 4:

0 ≤ u ≤ 2π

Sine equals 1/2 at:

u = π/6

and:

u = 5π/6

Solve for Time

First solution:

πt/2 = π/6

t = 1/3 s

Second solution:

πt/2 = 5π/6

t = 5/3 s

Final Answer

t = 1/3 s and 5/3 s

The object reaches a displacement of 2.5 cm twice during the first complete cycle.

Applications in Engineering: Rotating Machinery

A rotating component has angular position:

θ(t) = 120t

where θ is measured in degrees and t is measured in seconds.

A sensor activates whenever:

cos θ = 1/2

Find the activation times during the first 6 seconds.

Form the Equation

cos(120t) = 1/2

During the first 6 seconds:

0 ≤ t ≤ 6

Therefore:

0° ≤ 120t ≤ 720°

Find the Angles

Cosine equals 1/2 at:

60°, 300°, 420°, 660°

Solve for Time

120t = 60°

t = 0.5 s

120t = 300°

t = 2.5 s

120t = 420°

t = 3.5 s

120t = 660°

t = 5.5 s

Final Answer

t = 0.5 s, 2.5 s, 3.5 s, and 5.5 s

The sensor activates four times during the first 6 seconds.

Applications in Navigation and Geometry

Trigonometric equations are useful when determining directions and angles from known measurements.

Worked Example

A force has a horizontal component of 12 N and a magnitude of 20 N.

Find the possible directions relative to the positive horizontal axis for:

0° ≤ θ < 360°

Form the Equation

Fₓ = F cos θ

Substitute:

12 = 20cos θ

Divide:

cos θ = 12/20

cos θ = 0.6

Find the Reference Angle

θ = cos⁻¹(0.6)

θ ≈ 53.13°

Cosine is positive in Quadrants I and IV.

Therefore:

θ₁ ≈ 53.13°

θ₂ = 360° − 53.13°

θ₂ ≈ 306.87°

Final Answer

θ ≈ 53.13° or 306.87°

These represent two possible force directions with the same positive horizontal component.

A Strategy for Solving Trigonometric Equations

A systematic approach helps prevent missed or incorrect solutions.

Identify the Equation Type

Determine whether the equation involves:

• A single trigonometric function.

• An algebraic coefficient.

• A multiple angle.

• A phase shift.

• A trigonometric identity.

• A quadratic expression.

• Several trigonometric functions.

Simplify the Equation

Use algebra and identities to isolate or factor trigonometric expressions.

Find Reference Angles

Use exact special-angle values or inverse trigonometric functions.

Determine All Relevant Quadrants

Use the signs of sine, cosine, and tangent.

Account for Periodicity

Include all solutions in the specified interval.

Solve for the Original Variable

If the equation contains 2x, 3x, or x − 30°, remember to solve for x after finding the possible angle values.

Verify

Substitute solutions into the original equation.

State the Final Answer

List solutions in ascending order and include the appropriate units.

Common Mistakes

Finding only the principal value

Inverse trigonometric functions usually return one angle, but the equation may have several solutions.

Ignoring the specified interval

Only solutions within the given interval should be included.

Forgetting periodicity

Sine and cosine repeat every 360°, while tangent repeats every 180°.

Using incorrect quadrant signs

Check whether the trigonometric function is positive or negative in each quadrant.

Confusing degrees and radians

Use consistent angle units throughout the calculation.

Forgetting to adjust the interval for multiple angles

For sin 2x, the angle 2x covers twice the interval of x.

Dividing by a trigonometric expression that may equal zero

This can eliminate valid solutions.

Failing to factor quadratic expressions

Factoring often reveals multiple possible trigonometric values.

Accepting values outside the range of sine or cosine

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

Squaring without checking for extraneous solutions

Squaring can introduce values that do not satisfy the original equation.

Forgetting interval endpoints

An endpoint may be included or excluded depending on the inequality symbols.

Rounding too early

Keep sufficient precision until the final answer.

Not verifying solutions

Substitution into the original equation helps identify mistakes.

Key Terms

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

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

General Solution: A formula representing all solutions to a periodic equation.

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

Periodic Function: A function whose values repeat at regular intervals.

Period: The interval required for a function to complete one cycle.

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

Principal Value: The unique angle returned by an inverse trigonometric function within its defined range.

Inverse Sine: The function arcsin(x), used to determine an angle from its sine value.

Inverse Cosine: The function arccos(x), used to determine an angle from its cosine value.

Inverse Tangent: The function arctan(x), used to determine an angle from its tangent value.

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

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

Coterminal Angles: Angles that differ by complete revolutions and share the same terminal side.

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

Pythagorean Identity: A trigonometric relationship derived from sin² x + cos² x = 1.

Double-Angle Identity: An identity expressing a trigonometric function of 2x using functions of x.

Factoring: Rewriting an expression as a product of factors.

Quadratic Trigonometric Equation: A trigonometric equation containing a squared trigonometric function that can be treated like a quadratic equation.

Substitution: Replacing an expression with another variable or equivalent expression to simplify an equation.

Extraneous Solution: A value obtained during solving that does not satisfy the original equation.

Verification: Checking a proposed solution by substituting it into the original equation.

Graphical Solution: A solution found by identifying intersections of graphs.

Phase Shift: A horizontal translation of a periodic function.

Amplitude: The maximum displacement of a sinusoidal function from its midline.

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

Simple Harmonic Motion: Oscillatory motion that can be modelled using sine or cosine functions.

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

Check Your Understanding

1. Explain why trigonometric equations may have multiple solutions.

2. Solve sin x = 1/2 for 0° ≤ x < 360°.

3. Solve cos x = −√3/2 for 0° ≤ x < 360°.

4. Solve tan x = −1 for 0° ≤ x < 360°.

5. Solve sin x = √2/2 for 0 ≤ x < 2π.

6. Write the general solution of tan x = √3 in degrees.

7. Solve 3sin x − 2 = 0 for 0° ≤ x < 360°. Give answers to two decimal places.

8. Explain why cos x = 1.4 has no real solutions.

9. Solve sin 2x = √3/2 for 0° ≤ x < 360°.

10. Solve cos 3x = 1 for 0° ≤ x < 360°.

11. Solve sin(x − 45°) = √2/2 for 0° ≤ x < 360°.

12. Solve sin² x − sin x = 0 for 0° ≤ x < 360°.

13. Solve 2cos² x − 3cos x + 1 = 0 for 0° ≤ x < 360°.

14. Solve sin 2x = cos x for 0° ≤ x < 360°. Verify every solution.

15. A vibrating object has displacement x(t) = 8sin(πt/3), where x is measured in centimetres and t in seconds. Determine all times between 0 and 6 seconds when the displacement is 4 cm.

Key Takeaways

• Trigonometric equations involve unknown angles within trigonometric functions.

• Trigonometric equations often have multiple solutions because sine, cosine, and tangent are periodic.

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

• Tangent repeats every 180°, or π radians.

• Inverse trigonometric functions provide principal angles, but additional solutions may exist.

• Reference angles help identify solutions in different quadrants.

• The unit circle provides a visual method for locating trigonometric solutions.

• General solutions use an integer parameter to represent infinitely many angles.

• A specified interval determines which solutions must be included.

• Equations involving coefficients should first be rearranged to isolate the trigonometric function.

• Sine and cosine equations have no real solutions when the required function value lies outside [−1, 1].

• Multiple-angle equations require careful adjustment of the angle interval.

• Horizontal translations must be accounted for when solving equations involving phase shifts.

• Pythagorean and double-angle identities can simplify trigonometric equations.

• Factoring is useful for equations involving products and powers of trigonometric functions.

• Quadratic trigonometric equations can often be solved using substitution.

• Dividing by a trigonometric factor may eliminate valid solutions.

• Squaring equations can introduce extraneous solutions.

• Graphical methods can help estimate solutions and verify algebraic results.

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

• Trigonometric equations are used in wave motion, oscillations, rotating machinery, navigation, and vector analysis.

• Mastering trigonometric equations provides an essential foundation for advanced trigonometry, mathematical modelling, physics, engineering, and calculus.