Chapter 04 – Part 3: Trigonometric Equations and General Solutions.

CHAPTER 04: TRIGONOMETRY

PART 3: TRIGONOMETRIC EQUATIONS AND GENERAL SOLUTIONS

Introduction

In Part 1, we studied trigonometric functions, angles and graphs. In Part 2, we examined identities and formulae.

We now use this knowledge to solve trigonometric equations. Unlike ordinary algebraic equations, trigonometric equations can have several solutions because sine, cosine and tangent are periodic functions.

Our task is to find every solution in a specified interval and, where required, express the complete set of solutions using general solution formulae.

Learning Objectives

By the end of this lesson, you should be able to:

• Explain what a trigonometric equation is.
• Solve simple sine, cosine and tangent equations.
• Determine all solutions within a given interval.
• Write general solutions using integer multiples of the period.
• Solve trigonometric equations involving squared functions.
• Factorise quadratic trigonometric equations.
• Use identities to transform equations before solving.
• Check solutions and avoid losing valid answers.

1. Meaning of a Trigonometric Equation

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

Examples include:

sin θ = 1/2

cos θ = 0

tan θ = √3

2sin²θ − 3sin θ + 1 = 0

These equations are solved by finding the angle or angles that make the equation true.

Important:

Always read the interval carefully. An equation may have infinitely many solutions, but a specified interval contains only a limited number of them.

2. Understanding Periodicity

Trigonometric functions repeat their values at regular intervals. This repeating property is called periodicity.

Function	Period in degrees	Period in radians
sin θ	360°	2π
cos θ	360°	2π
tan θ	180°	π

For example:

sin 30° = sin 390° = sin 750°

Each angle differs from the preceding one by 360°.

Similarly:

tan 45° = tan 225° = tan 405°

Each angle differs by 180°.

Remember:

Sine and cosine repeat after 360°, whereas tangent repeats after 180°. This difference is essential when writing general solutions.

3. Solving Sine Equations

Consider the equation:

sin θ = 1/2

From the special angles, the reference angle is 30°.

Sine is positive in the first and second quadrants. Therefore, between 0° and 360°, the solutions are:

θ = 30° or θ = 150°

Because sine has a period of 360°, the general solutions are:

θ = 30° + 360°n
or
θ = 150° + 360°n

where n is any integer.

General Rule for Sine

If sin θ = sin α, the general solutions in degrees are:

θ = α + 360°n
or
θ = 180° − α + 360°n

For radians:

θ = α + 2πn
or
θ = π − α + 2πn

These forms give the two families of angles having the same sine value.

Worked Example 1

Solve sin θ = 1/2 for 0° ≤ θ ≤ 360°.

The reference angle is 30° because sin 30° = 1/2.

Sine is positive in Quadrants I and II.

Quadrant I: θ = 30°.
Quadrant II: θ = 180° − 30° = 150°.

Answer: θ = 30°, 150°.

Worked Example 2

Solve sin θ = −√3/2 for 0° ≤ θ ≤ 360°.

The reference angle is 60°.

Sine is negative in Quadrants III and IV.

Quadrant III: θ = 180° + 60° = 240°.
Quadrant IV: θ = 360° − 60° = 300°.

Answer: θ = 240°, 300°.

4. Solving Cosine Equations

Consider:

cos θ = 1/2

The reference angle is 60°.

Cosine is positive in Quadrants I and IV.

θ = 60° or θ = 300°

Since cosine has a period of 360°, the general solution is:

θ = α + 360°n
or
θ = 360° − α + 360°n

Equivalently:

θ = ±α + 360°n

For radians:

θ = ±α + 2πn

Worked Example 3

Solve cos θ = −1/2 for 0° ≤ θ ≤ 360°.

The reference angle is 60°.

Cosine is negative in Quadrants II and III.

Quadrant II: θ = 180° − 60° = 120°.
Quadrant III: θ = 180° + 60° = 240°.

Answer: θ = 120°, 240°.

5. Solving Tangent Equations

Consider:

tan θ = 1

The reference angle is 45°.

Tangent is positive in Quadrants I and III.

θ = 45° or θ = 225°

Since tangent has a period of 180°, its general solution is:

θ = α + 180°n

For radians:

θ = α + πn

Worked Example 4

Solve tan θ = −√3 for 0° ≤ θ ≤ 360°.

The reference angle is 60°.

Tangent is negative in Quadrants II and IV.

Quadrant II: θ = 180° − 60° = 120°.
Quadrant IV: θ = 360° − 60° = 300°.

Answer: θ = 120°, 300°.

6. Finding All Solutions in a Given Interval

When solving an equation over an interval larger than one complete revolution, more solutions may occur.

For example, consider:

Solve sin θ = 1/2 for 0° ≤ θ ≤ 720°.

The two families of solutions are:

θ = 30° + 360°n

θ = 150° + 360°n

For the first family:

• n = 0 gives 30°.
• n = 1 gives 390°.

For the second family:

• n = 0 gives 150°.
• n = 1 gives 510°.

Other integer values give angles outside the interval.

Answer:

θ = 30°, 150°, 390°, 510°

Worked Example 5

Solve cos θ = √2/2 for −360° ≤ θ ≤ 360°.

The reference angle is 45°.

Cosine is positive in Quadrants I and IV.

General solutions:
θ = 45° + 360°n
or θ = −45° + 360°n

First family:

• n = 0 gives 45°.
• n = −1 gives −315°.

Second family:

• n = 0 gives −45°.
• n = 1 gives 315°.

Answer:

θ = −315°, −45°, 45°, 315°

7. Equations Involving Squared Trigonometric Functions

Some trigonometric equations contain sin²θ, cos²θ or tan²θ. These can often be solved using algebraic methods.

For example:

2sin²θ − 1 = 0

Rearrange:

2sin²θ = 1

sin²θ = 1/2

Taking square roots:

sin θ = ±√2/2

We must solve both the positive and negative cases.

Worked Example 6

Solve 2sin²θ − 1 = 0 for 0° ≤ θ ≤ 360°.

2sin²θ − 1 = 0
2sin²θ = 1
sin²θ = 1/2

sin θ = ±√2/2

For sin θ = √2/2:

θ = 45°, 135°.

For sin θ = −√2/2:

θ = 225°, 315°.

Answer:

θ = 45°, 135°, 225°, 315°

8. Quadratic Trigonometric Equations

A quadratic trigonometric equation can be formed when a trigonometric function appears as the variable in a quadratic expression.

For example:

2sin²θ − 3sin θ + 1 = 0

We can solve this using factorisation in the same way as an ordinary quadratic equation.

Worked Example 7

Solve 2sin²θ − 3sin θ + 1 = 0 for 0° ≤ θ ≤ 360°.

Factorise:
(2sin θ − 1)(sin θ − 1) = 0

Therefore:
2sin θ − 1 = 0
or
sin θ − 1 = 0

First equation:
sin θ = 1/2
θ = 30°, 150°.

Second equation:
sin θ = 1
θ = 90°.

Answer:

θ = 30°, 90°, 150°

Important examination warning

Never cancel a trigonometric factor before considering whether it can equal zero.

For example, if:

cos θ(sin θ + 1) = 0

the correct method is to set each factor equal to zero:

cos θ = 0

or sin θ + 1 = 0

Cancelling cos θ would lose the solutions obtained from cos θ = 0.

Worked Example 8

Solve cos θ(sin θ + 1) = 0 for 0° ≤ θ ≤ 360°.

Set each factor equal to zero.
cos θ = 0
or sin θ + 1 = 0.

From cos θ = 0:
θ = 90°, 270°.

From sin θ = −1:
θ = 270°.

Combining the distinct solutions:

Answer: θ = 90°, 270°.

9. Using Pythagorean Identities in Equations

Sometimes an equation contains more than one trigonometric function. An identity can help convert it into a simpler equation.

Recall:

sin²θ + cos²θ = 1

Therefore:

cos²θ = 1 − sin²θ

We can use this substitution when an equation contains both sine and cosine.

Worked Example 9

Solve 2cos²θ = 1 for 0° ≤ θ ≤ 360°.

2cos²θ = 1
cos²θ = 1/2

cos θ = ±√2/2

For cos θ = √2/2:
θ = 45°, 315°.

For cos θ = −√2/2:
θ = 135°, 225°.

Answer:

θ = 45°, 135°, 225°, 315°

10. Equations in Radians

Trigonometric equations may also be expressed in radians. The same methods apply, but the periods are written in terms of π.

Function	Period
sin θ	2π
cos θ	2π
tan θ	π

Worked Example 10

Solve sin θ = 1/2 for 0 ≤ θ ≤ 2π.

The reference angle is π/6.

Sine is positive in Quadrants I and II.

Quadrant I:
θ = π/6.

Quadrant II:
θ = π − π/6 = 5π/6.

Answer:

θ = π/6, 5π/6

11. Common Errors to Avoid

• Giving only one solution when the equation has more than one solution in the interval.
• Using 360° as the period of tangent instead of 180°.
• Forgetting the negative square-root case when solving sin²θ or cos²θ equations.
• Ignoring the interval stated in the question.
• Using the wrong quadrants for positive and negative trigonometric values.
• Cancelling a trigonometric factor that could be zero.
• Mixing degrees and radians in the same calculation.
• Repeating a solution or including an angle outside the specified interval.

12. Practice Exercises

Exercise A: Basic Equations

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

2. Solve cos θ = −1/2 for 0° ≤ θ ≤ 360°.

3. Solve tan θ = √3 for 0° ≤ θ ≤ 360°.

4. Solve sin θ = −1/2 for 0° ≤ θ ≤ 360°.

5. Solve cos θ = √3/2 for 0° ≤ θ ≤ 360°.

6. Solve tan θ = −1 for 0° ≤ θ ≤ 360°.

Exercise B: Extended Intervals

7. Solve sin θ = 1/2 for 0° ≤ θ ≤ 720°.

8. Solve cos θ = 1/2 for −360° ≤ θ ≤ 360°.

9. Solve tan θ = 1 for −180° ≤ θ ≤ 540°.

10. Solve sin θ = −√3/2 for 0° ≤ θ ≤ 720°.

Exercise C: Quadratic Equations

11. Solve 2sin²θ − 1 = 0 for 0° ≤ θ ≤ 360°.

12. Solve 2cos²θ − 1 = 0 for 0° ≤ θ ≤ 360°.

13. Solve 2sin²θ − 3sin θ + 1 = 0 for 0° ≤ θ ≤ 360°.

14. Solve 2cos²θ + cos θ − 1 = 0 for 0° ≤ θ ≤ 360°.

15. Solve sin θ(2sin θ − 1) = 0 for 0° ≤ θ ≤ 360°.

Exercise D: Mixed Equations

16. Solve cos θ(sin θ + 1) = 0 for 0° ≤ θ ≤ 360°.

17. Solve sin²θ = cos²θ for 0° ≤ θ ≤ 360°.

18. Solve 1 + tan²θ = 4 for 0° ≤ θ ≤ 360°.

19. Solve sin θ = 1/2 for 0 ≤ θ ≤ 2π.

20. Solve cos θ = −1/2 for 0 ≤ θ ≤ 2π.

13. Answers and Worked Solutions

Answers to Exercise A

1. Reference angle = 30°. Sine is positive in Quadrants I and II.

θ = 30°, 150°.

2. Reference angle = 60°. Cosine is negative in Quadrants II and III.

θ = 120°, 240°.

3. Reference angle = 60°. Tangent is positive in Quadrants I and III.

θ = 60°, 240°.

4. Reference angle = 30°. Sine is negative in Quadrants III and IV.

θ = 210°, 330°.

5. Reference angle = 30°. Cosine is positive in Quadrants I and IV.

θ = 30°, 330°.

6. Reference angle = 45°. Tangent is negative in Quadrants II and IV.

θ = 135°, 315°.

Worked solutions to Exercise B

7.

sin θ = 1/2.

General solutions:

θ = 30° + 360°n or θ = 150° + 360°n.

Within the interval:

θ = 30°, 150°, 390°, 510°.

8.

cos θ = 1/2.

General solutions:

θ = 60° + 360°n or θ = −60° + 360°n.

Within the interval:

θ = −300°, −60°, 60°, 300°.

9.

tan θ = 1.

General solution:

θ = 45° + 180°n.

Within the interval:

θ = −135°, 45°, 225°, 405°.

10.

Reference angle = 60°.

General solutions:

θ = 240° + 360°n or θ = 300° + 360°n.

Within the interval:

θ = 240°, 300°, 600°, 660°.

Worked solutions to Exercise C

11.

2sin²θ − 1 = 0

sin²θ = 1/2

sin θ = ±√2/2.

θ = 45°, 135°, 225°, 315°.

12.

2cos²θ − 1 = 0

cos²θ = 1/2

cos θ = ±√2/2.

θ = 45°, 135°, 225°, 315°.

13.

2sin²θ − 3sin θ + 1 = 0

(2sin θ − 1)(sin θ − 1) = 0.

sin θ = 1/2 or sin θ = 1.

θ = 30°, 90°, 150°.

14.

2cos²θ + cos θ − 1 = 0

(2cos θ − 1)(cos θ + 1) = 0.

cos θ = 1/2 or cos θ = −1.

θ = 60°, 180°, 300°.

15.

sin θ(2sin θ − 1) = 0.

sin θ = 0 or sin θ = 1/2.

From sin θ = 0: θ = 0°, 180°, 360°.

From sin θ = 1/2: θ = 30°, 150°.

θ = 0°, 30°, 150°, 180°, 360°.

Worked solutions to Exercise D

16.

cos θ(sin θ + 1) = 0.

cos θ = 0 or sin θ = −1.

From cos θ = 0: θ = 90°, 270°.

From sin θ = −1: θ = 270°.

θ = 90°, 270°.

17.

sin²θ = cos²θ.

Dividing by cos²θ, where cos θ ≠ 0:

tan²θ = 1.

tan θ = ±1.

Checking the original equation gives:

θ = 45°, 135°, 225°, 315°.

18.

1 + tan²θ = 4.

tan²θ = 3.

tan θ = ±√3.

θ = 60°, 120°, 240°, 300°.

19.

sin θ = 1/2.

In the interval 0 to 2π:

θ = π/6, 5π/6.

20.

cos θ = −1/2.

In the interval 0 to 2π:

θ = 2π/3, 4π/3.

14. Final Revision Summary

Sine general solution
θ = α + 360°n
or θ = 180° − α + 360°n

Cosine general solution
θ = ±α + 360°n

Tangent general solution
θ = α + 180°n

For radians, replace 360° with 2π and 180° with π.

Here, α is a reference angle appropriate to the equation, and n is any integer. Always select the values that fall within the required interval.

Final examination checklist

• Identify the trigonometric function.
• Find the reference angle.
• Determine the correct quadrants.
• Write all possible solutions.
• Apply the interval limits.
• Check that no valid solutions have been lost.

Looking Ahead: Part 4

In the next lesson, we shall study Compound Angles, Double Angles and Multiple Angles. We will develop the formulae for sin(A ± B), cos(A ± B) and tan(A ± B), and learn how to apply them in mathematical problems.

MATHEMATICS HIP-HOP
Making Mathematics Easier, One Concept at a Time.
A-Level Pure Mathematics — Chapter 04, Part 3