Open Math›Year 11›Mathematics Advanced›Trigonometric identities and equations›Trigonometric equations

Trigonometric equations

Solve trigonometric equations for all solutions in a given domain, including equations requiring Pythagorean identity substitution and equations that reduce to a quadratic in a trigonometric function.

Worked examples

Solving a basic trigonometric equation

Straightforward

Problem

Solve

\cos\theta = \dfrac{1}{2}

for

0° \leq \theta \leq 360°

Find the reference angle.

\theta_r = \cos^{-1}\left(\frac{1}{2}\right) = 60°

Identify the quadrants where cosine is positive.

Cosine is positive in the first and fourth quadrants (Q1 and Q4).

Write the solution in the first quadrant.

\theta = 60°

Write the solution in the fourth quadrant.

\theta = 360° - 60° = 300°

Answer

\theta = 60°

and

\theta = 300°

Solving using a Pythagorean identity substitution

Moderate

Problem

Solve

2\cos^2\theta + 5\sin\theta + 1 = 0

for

0° \leq \theta \leq 360°

Replace

\cos^2\theta

using the identity

\cos^2\theta = 1 - \sin^2\theta

2(1 - \sin^2\theta) + 5\sin\theta + 1 = 0

Expand and collect terms.

2 - 2\sin^2\theta + 5\sin\theta + 1 = 0

-2\sin^2\theta + 5\sin\theta + 3 = 0

2\sin^2\theta - 5\sin\theta - 3 = 0

Factorise the quadratic in

\sin\theta

(2\sin\theta + 1)(\sin\theta - 3) = 0

Solve each factor. Reject any value outside

[-1, 1]

\sin\theta = -\dfrac{1}{2}

\sin\theta = 3

(impossible, since

-1 \leq \sin\theta \leq 1

)

Find all angles where

\sin\theta = -\dfrac{1}{2}

[0°, 360°]

. Sine is negative in Q3 and Q4.

Reference angle:

\sin^{-1}\left(\dfrac{1}{2}\right) = 30°

\theta = 180° + 30° = 210° \quad \text{and} \quad \theta = 360° - 30° = 330°

Answer

\theta = 210°

and

\theta = 330°

Solving a quadratic trigonometric equation

Challenging

Problem

Solve

2\cos^2\theta - \cos\theta - 1 = 0

for

0 \leq \theta \leq 2\pi

Recognise the equation as a quadratic in

\cos\theta

. Let

u = \cos\theta

2u^2 - u - 1 = 0

Factorise.

(2u + 1)(u - 1) = 0

Substitute back and state the two equations to solve.

\cos\theta = -\frac{1}{2} \quad \text{or} \quad \cos\theta = 1

Find all angles where

\cos\theta = -\dfrac{1}{2}

[0, 2\pi]

. Cosine is negative in Q2 and Q3.

Reference angle:

\cos^{-1}\left(\dfrac{1}{2}\right) = \dfrac{\pi}{3}

\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \quad \text{and} \quad \theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}

Find all angles where

\cos\theta = 1

[0, 2\pi]

\theta = 0 \quad \text{and} \quad \theta = 2\pi

Answer

\theta = 0, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}, 2\pi

Practise

Q1·Straightforward

Solve

\sin\theta = \dfrac{1}{2}

for

0° \leq \theta \leq 360°

. Which list contains all solutions?

$30°$ only$30°$ and $150°$$30°$ and $330°$$150°$ and $210°$

Explanation

The reference angle is

\sin^{-1}\left(\dfrac{1}{2}\right) = 30°

.

Since

\sin\theta > 0

, solutions lie in the first and second quadrants:

\theta = 30° \quad \text{and} \quad \theta = 180° - 30° = 150°

Q2·Straightforward

Solve

\cos\theta = 0

for

0° \leq \theta \leq 360°

. The smaller solution is

90°

. What is the larger solution in degrees?

Explanation

Cosine is zero when

\theta

lies on the

y

-axis of the unit circle.

In the interval

0° \leq \theta \leq 360°

, the solutions are:

\theta = 90° \quad \text{and} \quad \theta = 270°

The larger solution is

270°

Q3·Straightforward

Solve

\tan\theta = \sqrt{3}

for

0° \leq \theta \leq 360°

. The smaller solution is

60°

. What is the larger solution in degrees?

Explanation

The reference angle is

\tan^{-1}(\sqrt{3}) = 60°

.

Since

\tan\theta > 0

, solutions lie in the first and third quadrants:

\theta = 60° \quad \text{and} \quad \theta = 180° + 60° = 240°

The larger solution is

240°

Q4·Straightforward

Solve

\sin\theta = -1

for

0 \leq \theta \leq 2\pi

. What is the solution?

$0$$\dfrac{\pi}{2}$$\pi$$\dfrac{3\pi}{2}$

Explanation

Sine reaches its minimum value of

-1

at the bottom of the unit circle, which corresponds to

\theta = \dfrac{3\pi}{2}

.

Verification:

\sin\left(\dfrac{3\pi}{2}\right) = -1

\checkmark

Q5·Moderate

Solve

2\cos^2\theta + 3\sin\theta = 3

for

0° \leq \theta \leq 360°

. What is the largest solution in degrees?

Explanation

Replace

\cos^2\theta

with

1 - \sin^2\theta

2(1 - \sin^2\theta) + 3\sin\theta = 3

2 - 2\sin^2\theta + 3\sin\theta = 3

2\sin^2\theta - 3\sin\theta + 1 = 0

(2\sin\theta - 1)(\sin\theta - 1) = 0

\sin\theta = \dfrac{1}{2}

\theta = 30°, 150°

\sin\theta = 1

\theta = 90°

The three solutions are

30°, 90°, 150°

. The largest is

\mathbf{150°}

Q6·Moderate

Solve

\cos^2\theta = \dfrac{3}{2}\sin\theta

for

0° \leq \theta \leq 360°

. What is the sum of all solutions in degrees?

Explanation

Replace

\cos^2\theta

with

1 - \sin^2\theta

1 - \sin^2\theta = \frac{3}{2}\sin\theta

Multiply both sides by 2:

2 - 2\sin^2\theta = 3\sin\theta

2\sin^2\theta + 3\sin\theta - 2 = 0

(2\sin\theta - 1)(\sin\theta + 2) = 0

\sin\theta = \dfrac{1}{2}

\theta = 30°, 150°

\sin\theta = -2

: impossible (since

-1 \leq \sin\theta \leq 1

)

Sum of solutions

= 30° + 150° = 180°

Q7·Moderate

Solve

2\sin^2\theta - \cos\theta - 1 = 0

for

0° \leq \theta \leq 360°

. What is the sum of all solutions in degrees?

Explanation

Replace

\sin^2\theta

with

1 - \cos^2\theta

2(1 - \cos^2\theta) - \cos\theta - 1 = 0

2 - 2\cos^2\theta - \cos\theta - 1 = 0

2\cos^2\theta + \cos\theta - 1 = 0

(2\cos\theta - 1)(\cos\theta + 1) = 0

\cos\theta = \dfrac{1}{2}

\theta = 60°, 300°

\cos\theta = -1

\theta = 180°

Sum of solutions

= 60° + 180° + 300° = 540°

Q8·Moderate

Solve

2\cos^2\theta - \sin\theta - 1 = 0

for

0° \leq \theta \leq 360°

. What is the sum of all solutions in degrees?

Explanation

Replace

\cos^2\theta

with

1 - \sin^2\theta

2(1 - \sin^2\theta) - \sin\theta - 1 = 0

2 - 2\sin^2\theta - \sin\theta - 1 = 0

2\sin^2\theta + \sin\theta - 1 = 0

(2\sin\theta - 1)(\sin\theta + 1) = 0

\sin\theta = \dfrac{1}{2}

\theta = 30°, 150°

\sin\theta = -1

\theta = 270°

Sum of solutions

= 30° + 150° + 270° = 450°

Q9·Challenging

Solve

2\sin^2\theta - \sin\theta - 1 = 0

for

0 \leq \theta \leq 2\pi

. Which set contains all solutions?

$\dfrac{\pi}{2}$ only$\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}$$\dfrac{\pi}{2}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$$\dfrac{\pi}{2}, \dfrac{5\pi}{6}, \dfrac{11\pi}{6}$

Explanation

Let

u = \sin\theta

2u^2 - u - 1 = 0

(2u + 1)(u - 1) = 0

\sin\theta = -\dfrac{1}{2}

: sine is negative in Q3 and Q4. Reference angle

= \dfrac{\pi}{6}

\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \quad \text{and} \quad \theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}

\sin\theta = 1

\theta = \dfrac{\pi}{2}

Solutions:

\dfrac{\pi}{2}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}

Q10·Challenging

Solve

\tan^2\theta - \tan\theta = 0

for

0 \leq \theta < 2\pi

. Which set contains all solutions?

$0, \dfrac{\pi}{4}, \pi, \dfrac{5\pi}{4}$$\dfrac{\pi}{4}, \pi, \dfrac{5\pi}{4}, \dfrac{7\pi}{4}$$0, \dfrac{\pi}{4}, \dfrac{\pi}{2}, \dfrac{5\pi}{4}$$0, \pi, \dfrac{\pi}{4}, \dfrac{3\pi}{4}$

Explanation

Factor:

\tan\theta(\tan\theta - 1) = 0

\tan\theta = 0

\theta = 0

and

\theta = \pi

\tan\theta = 1

: reference angle

\dfrac{\pi}{4}

; tangent is positive in Q1 and Q3.

\theta = \frac{\pi}{4} \quad \text{and} \quad \theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4}

Solutions:

0, \dfrac{\pi}{4}, \pi, \dfrac{5\pi}{4}

Q11·Challenging

Solve

4\sin^2\theta - 1 = 0

for

0 \leq \theta \leq 2\pi

. Which set contains all solutions?

$\dfrac{\pi}{6}, \dfrac{5\pi}{6}$$\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}$$\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$$\dfrac{\pi}{3}, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}, \dfrac{5\pi}{3}$

Explanation

Factorise using difference of two squares:

4\sin^2\theta - 1 = (2\sin\theta - 1)(2\sin\theta + 1) = 0

\sin\theta = \dfrac{1}{2}

: sine positive in Q1 and Q2. Reference angle

= \dfrac{\pi}{6}

\theta = \frac{\pi}{6} \quad \text{and} \quad \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}

\sin\theta = -\dfrac{1}{2}

: sine negative in Q3 and Q4.

\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \quad \text{and} \quad \theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}

Solutions:

\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}

Q12·Challenging

Solve

\cos^2\theta + \cos\theta = 0

for

0 \leq \theta \leq 2\pi

. Which set contains all solutions?

$\dfrac{\pi}{2}, \pi$$\pi, \dfrac{3\pi}{2}$$\dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2}$$0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2}$

Explanation

Factor:

\cos\theta(\cos\theta + 1) = 0

\cos\theta = 0

\theta = \dfrac{\pi}{2}

and

\theta = \dfrac{3\pi}{2}

\cos\theta = -1

\theta = \pi

Solutions:

\dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2}