Solving Trigonometric Equations — Edexcel A-level Mathematics (Pure)

SPEC 5.7 • TRIGONOMETRY

Lesson 5.12A: Solving Trigonometric Equations

Basic equations, any angles, and factorising

Basic Equations & CAST

To solve equations like

sin θ = k

,

cos θ = k

, or

tan θ = k

:

Method

Find the principal value (acute angle) using inverse trig

Use CAST to identify which quadrants have solutions

Draw lines in those quadrants at the acute angle to the horizontal axis

Read off all angles in the given range

Measuring Angles

Always measure from the nearest horizontal line (0° or 180°):

Anticlockwise = positive

Clockwise = negative

Important

Never measure from the vertical axis (90° or 270°). Always use horizontal lines only.

Worked Example01

Solve $sin θ = 0.5$ for $0° \leq θ \leq 360°$ .

1

Principal value:

θ = sin^{- 1} (0.5) = 30°

2

Sin is positive → quadrants 1 (A) and 2 (S)

3

Draw lines at 30° to x-axis in Q1 and Q2

4

Q1: 30° anticlockwise from 0° →

θ = 30°

5

Q2: 30° anticlockwise from 180° →

θ = 150°

Answerθ = 30° and θ = 150°

Worked Example02

Solve $2 cos x + 1 = 0$ for $- π \leq x \leq π$ .

1

Rearrange:

2 cos x = - 1

→

cos x = - \frac{1}{2}

2

Reference angle:

cos^{- 1} (\frac{1}{2}) = \frac{π}{3}

3

Cos is negative → quadrants 2 (S) and 3 (T)

4

Q2:

\frac{π}{3}

anticlockwise from π →

x = \frac{2 π}{3}

5

Q3:

\frac{π}{3}

clockwise from −π →

x = - \frac{2 π}{3}

Answer

x = - \frac{2 π}{3}

and

x = \frac{2 π}{3}

Equations with Any Angle

For equations where the angle is not just θ (e.g., 2θ, 3x, θ + 30°), use a substitution.

Method

Let A = (the angle expression)

Adjust the range — apply the same operation to the range limits

Solve for A, finding ALL solutions in the new range

Reverse the substitution to find θ (or x)

Examples of Range Adjustment

SubstitutionOriginal RangeNew Range for AA = 2θ0° ≤ θ ≤ 360°0° ≤ A ≤ 720°A = 3x−π ≤ x ≤ π−3π ≤ A ≤ 3πA = θ − 30°0° ≤ θ ≤ 360°−30° ≤ A ≤ 330°

Worked Example03

Solve $sin 2 θ = 0.5$ for $0° \leq θ \leq 360°$ .

1

Let

A = 2 θ

. Range:

0° \leq A \leq 720°

2

Solve

sin A = 0.5

. Principal value:

A = 30°

3

Sin positive → Q1 (A) and Q2 (S)

4

First cycle:

A = 30°, 150°

5

Second cycle (+360°):

A = 390°, 510°

6

Reverse:

θ = \frac{A}{2}

→

θ = 15°, 75°, 195°, 255°

Answerθ = 15°, 75°, 195°, 255°

Worked Example04

Solve $cos (x + \frac{π}{4}) = - \frac{2}{2}$ for $- π \leq x \leq π$ .

1

Let

A = x + \frac{π}{4}

. Range:

- \frac{3 π}{4} \leq A \leq \frac{5 π}{4}

2

Reference angle:

cos^{- 1} (\frac{2}{2}) = \frac{π}{4}

3

Cos negative → Q2 (S) and Q3 (T)

4

Q2:

\frac{π}{4}

anticlockwise from π →

A = \frac{3 π}{4}

(correct)

5

Q3:

\frac{π}{4}

clockwise from −π →

A = - \frac{3 π}{4}

(correct)

6

Reverse:

x = A - \frac{π}{4}

→

x = \frac{π}{2}, - π

Answer

x = - π

and

x = \frac{π}{2}

Factorising Equations

Some trigonometric equations can be solved by factorising, just like algebraic equations.

Method

Rearrange so one side equals zero

Factorise — look for common factors or quadratic patterns

Set each factor equal to zero

Solve each resulting equation using CAST

Never Divide by Trig Functions!

If you see

sin θ \cdot cos θ = sin θ

, don't divide by sin θ — you'll lose solutions where sin θ = 0! Instead, rearrange to

sin θ (cos θ - 1) = 0

and factorise.

Worked Example05

Solve $sin θ (2 sin θ - 1) = 0$ for $0° \leq θ \leq 360°$ .

1

Already factorised. Set each factor = 0

2

sin θ = 0

:

θ = 0°, 180°, 360°

3

2 sin θ - 1 = 0

→

sin θ = \frac{1}{2}

4

Sin positive → Q1 (A) and Q2 (S)

5

θ = 30°, 150°

Answerθ = 0°, 30°, 150°, 180°, 360°

Worked Example06

Solve $2 sin^{2} x - 3 sin x + 1 = 0$ for $0 \leq x \leq 2 π$ .

1

Factorise:

(2 sin x - 1) (sin x - 1) = 0

2

2 sin x - 1 = 0

→

sin x = \frac{1}{2}

3

Sin positive → Q1 (A) and Q2 (S)

4

x = \frac{π}{6}, \frac{5 π}{6}

5

sin x - 1 = 0

→

sin x = 1

→

x = \frac{π}{2}

Answer

x = \frac{π}{6}, \frac{π}{2}, \frac{5 π}{6}

Practice

Question: Solve

tan θ = - 1

for

- 180° \leq θ \leq 180°

.

1

Reference angle:

tan^{- 1} (1) = 45°

2

Tan is negative → quadrants 2 (S) and 4 (C)

3

Q2: 45° anticlockwise from 180° →

θ = 135°

4

Q4: 45° clockwise from 0° →

θ = - 45°

Answerθ = −45° and θ = 135°

Practice

Question: Solve

tan 2 x = 3

for

0 \leq x \leq π

.

1

Let

A = 2 x

. Range:

0 \leq A \leq 2 π

2

Solve

tan A = 3

. Principal value:

A = \frac{π}{3}

3

Tan positive → Q1 (A) and Q3 (T)

4

A = \frac{π}{3}, \frac{π}{3} + π = \frac{4 π}{3}

5

Reverse:

x = \frac{A}{2}

→

x = \frac{π}{6}, \frac{2 π}{3}

Answer

x = \frac{π}{6}

and

x = \frac{2 π}{3}

Practice

Question: Solve

tan^{2} θ - tan θ = 0

for

- 180° \leq θ \leq 180°

.

1

Factorise:

tan θ (tan θ - 1) = 0

2

tan θ = 0

:

θ = - 180°, 0°, 180°

3

tan θ = 1

: Tan positive → Q1 (A) and Q3 (T)

4

Q1: 45° anticlockwise from 0° →

θ = 45°

5

Q3: 45° clockwise from −180° →

θ = - 135°

Answerθ = −180°, −135°, 0°, 45°, 180°

Summary

Three Methods Covered

Equation TypeMethod

sin θ = k

,

cos θ = k

,

tan θ = k

Find principal value, use CAST, draw lines at acute angle

sin (n θ) = k

,

cos (x + a) = k

, etc.Let A = (angle), adjust range, solve, reverse substitution

sin θ (sin θ - 1) = 0

, quadraticsFactorise, set each factor = 0, solve separately

Key Reminders

Measure from the nearest horizontal line: anticlockwise = positive, clockwise = negative

For negative ranges, Q3 and Q4 give negative angles

When adjusting range, apply the same operation to both limits

Never divide by a trig function — factorise instead