Solving Trigonometric Equations (Pure Mathematics 1)

Introduction

Trigonometric equations appear in almost every 9709 Pure Mathematics 1 paper. Unlike finding a single angle on a calculator, the exam requires you to find all solutions lying within a given interval — typically $0° \le \theta \le 360°$ , $0 \le \theta \le 2\pi$ , or a shifted/scaled interval such as $-\pi \le \theta \le \pi$ . The syllabus requires you to solve simple equations (no general formula needed), but the method must be systematic and rigorous to earn full marks.

Core Concept

The CAST Diagram and Symmetry

Every trigonometric equation of the form $\sin\theta = k$ , $\cos\theta = k$ , or $\tan\theta = k$ has a principal value — the value your calculator returns. From this principal value, you use the symmetry of the trigonometric functions to find all other solutions in the required interval.

The CAST diagram summarises which ratios are positive in each quadrant:

Quadrant	Angles (degrees)	Angles (radians)	Positive ratio
1st (Q1)	$0°$ to $90°$	$0$ to $\tfrac{\pi}{2}$	All (sin, cos, tan)
2nd (Q2)	$90°$ to $180°$	$\tfrac{\pi}{2}$ to $\pi$	Sin only
3rd (Q3)	$180°$ to $270°$	$\pi$ to $\tfrac{3\pi}{2}$	Tan only
4th (Q4)	$270°$ to $360°$	$\tfrac{3\pi}{2}$ to $2\pi$	Cos only

Symmetry Rules

Let $\alpha$ denote the principal value (always taken as a positive acute angle from your calculator, i.e. $\alpha = |\text{calculator value}|$ ).

Equation	Solutions in $[0°, 360°]$
$\sin\theta = +k$	$\theta = \alpha$ and $\theta = 180° - \alpha$
$\sin\theta = -k$	$\theta = 180° + \alpha$ and $\theta = 360° - \alpha$
$\cos\theta = +k$	$\theta = \alpha$ and $\theta = 360° - \alpha$
$\cos\theta = -k$	$\theta = 180° - \alpha$ and $\theta = 180° + \alpha$
$\tan\theta = +k$	$\theta = \alpha$ and $\theta = 180° + \alpha$
$\tan\theta = -k$	$\theta = 180° - \alpha$ and $\theta = 360° - \alpha$

The same logic applies in radians, replacing $180°$ with $\pi$ and $360°$ with $2\pi$ .

Equations with a Transformed Angle

When the equation involves $\sin(2\theta)$ , $\cos(\theta + 30°)$ , etc., substitute $u = 2\theta$ (or the relevant expression), expand the interval for $u$ , solve for $u$ , then convert back to $\theta$ .

Key Formulae & Definitions

Pythagorean identity (often needed to reduce equations):

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

Derived forms:

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

Tangent identity (used to simplify mixed $\sin/\cos$ equations):

\tan\theta = \frac{\sin\theta}{\cos\theta}

Period reminders:

Function	Period (degrees)	Period (radians)
$\sin\theta$	$360°$	$2\pi$
$\cos\theta$	$360°$	$2\pi$
$\tan\theta$	$180°$	$\pi$

Worked Examples

Example 1 — Basic equation with a double angle

Solve $\sin 2\theta = \dfrac{\sqrt{3}}{2}$ for $0° \le \theta \le 360°$ .

Step 1 — Substitute. Let $u = 2\theta$ . The interval for $\theta$ is $[0°, 360°]$ , so the interval for $u$ is $[0°, 720°]$ .

Step 2 — Find the principal value. $\sin u = \dfrac{\sqrt{3}}{2}$ , so $\alpha = 60°$ (exact value).

Step 3 — List all solutions for $u$ in $[0°, 720°]$ . Since $\sin u > 0$ , solutions lie in Q1 and Q2:

u = 60°,\quad 120°,\quad 420°,\quad 480°

Step 4 — Convert back. $\theta = \dfrac{u}{2}$ :

\theta = 30°,\quad 60°,\quad 210°,\quad 240°

Answer: $\theta = 30°,\ 60°,\ 210°,\ 240°$

Example 2 — Quadratic in $\cos\theta$ using an identity

Solve $2\cos^2\theta + \cos\theta - 1 = 0$ for $0 \le \theta \le 2\pi$ . Give answers in radians.

Step 1 — Factorise. Treat $\cos\theta$ as the variable:

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

Step 2 — Solve each factor.

Factor 1: $2\cos\theta - 1 = 0 \Rightarrow \cos\theta = \dfrac{1}{2}$

Principal value: $\alpha = \dfrac{\pi}{3}$ . Since $\cos\theta > 0$ : solutions in Q1 and Q4:

\theta = \frac{\pi}{3}, \quad \theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}

Factor 2: $\cos\theta + 1 = 0 \Rightarrow \cos\theta = -1$

This occurs at the boundary: $\theta = \pi$

Step 3 — Collect and order all solutions.

\theta = \frac{\pi}{3},\quad \pi,\quad \frac{5\pi}{3}

Example 3 — Mixed equation reduced using $\tan$

Solve $3\sin\theta = 2\cos\theta$ for $0° \le \theta \le 360°$ .

Step 1 — Rearrange. Divide both sides by $\cos\theta$ (valid since $\cos\theta = 0$ gives $0 = 2$ , no solution):

\tan\theta = \frac{2}{3}

Step 2 — Principal value. $\alpha = \tan^{-1}\!\left(\dfrac{2}{3}\right) = 33.69°$ (to 2 d.p.)

Step 3 — Solutions in $[0°, 360°]$ . Since $\tan\theta > 0$ : Q1 and Q3:

\theta = 33.7°,\quad \theta = 180° + 33.69° = 213.7°

Answer: $\theta = 33.7°,\ 213.7°$ (to 1 d.p.)

Common Mistakes & Examiner Pitfalls

Forgetting to expand the interval. When solving $\sin(2\theta) = k$ , candidates solve only in $[0°, 360°]$ instead of $[0°, 720°]$ and miss half the solutions. Always multiply the interval endpoints by the coefficient of $\theta$ .
Using the calculator value directly as both answers. The calculator gives only one solution. You must always apply the symmetry rule to find the second (and further) solution(s) in the interval.
Sign errors with negative values. For $\sin\theta = -0.5$ , the solutions are in Q3 and Q4 — not Q1 and Q2. Always determine the quadrants using CAST after noting the sign of $k$ .
Dividing by a trig function without checking for extra solutions. If you divide by $\sin\theta$ or $\cos\theta$ , you may lose solutions where that function equals zero. Factorise instead wherever possible.
Mixing degrees and radians. If the interval is given in radians, all solutions must be in radians. Never mix the two in the same answer.
Not checking boundary values. Values like $\theta = 0$ , $\pi$ , $2\pi$ are valid solutions if they satisfy the equation — don't discard them.
Rounding prematurely. Keep the principal value to sufficient decimal places (or exact form) before applying symmetry rules; round only the final answers.

Practice Questions

Q1. Solve $\cos\theta = -\dfrac{1}{\sqrt{2}}$ for $0° \le \theta \le 360°$ .

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Principal value: $\alpha = \cos^{-1}\!\left(\dfrac{1}{\sqrt{2}}\right) = 45°$

Since $\cos\theta < 0$ : solutions in Q2 and Q3.

\theta = 180° - 45° = 135°, \qquad \theta = 180° + 45° = 225°

Answer: $\theta = 135°,\ 225°$

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Q2. Solve $\tan\!\left(\theta - 30°\right) = 1$ for $0° \le \theta \le 360°$ .

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Let $u = \theta - 30°$ . Interval for $u$ : $-30° \le u \le 330°$ .

$\tan u = 1$ , so $\alpha = 45°$ . Period of tan is $180°$ .

Solutions for $u$ in $[-30°, 330°]$ :

u = 45°, \qquad u = 45° + 180° = 225°

(Note: $u = 45° - 180° = -135°$ is outside the interval.)

Convert back: $\theta = u + 30°$ :

\theta = 75°, \qquad \theta = 255°

Answer: $\theta = 75°,\ 255°$

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Q3. Solve $2\sin^2\theta - \sin\theta - 1 = 0$ for $0 \le \theta \le 2\pi$ . Give exact answers.

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Factorise: $(2\sin\theta + 1)(\sin\theta - 1) = 0$

Factor 1: $\sin\theta = -\dfrac{1}{2}$ , so $\alpha = \dfrac{\pi}{6}$ . Solutions in Q3 and Q4:

\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}, \qquad \theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}

Factor 2: $\sin\theta = 1 \Rightarrow \theta = \dfrac{\pi}{2}$

Answer: $\theta = \dfrac{\pi}{2},\ \dfrac{7\pi}{6},\ \dfrac{11\pi}{6}$

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Q4. Solve $\sin 2\theta = -\cos 2\theta$ for $0° \le \theta \le 180°$ .

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Rearrange: $\dfrac{\sin 2\theta}{\cos 2\theta} = -1$ , so $\tan 2\theta = -1$ .

Let $u = 2\theta$ . Interval for $u$ : $[0°, 360°]$ .

$\alpha = 45°$ . Since $\tan u < 0$ : solutions in Q2 and Q4:

u = 180° - 45° = 135°, \qquad u = 360° - 45° = 315°

Convert: $\theta = \dfrac{u}{2}$ :

\theta = 67.5°, \qquad \theta = 157.5°

Answer: $\theta = 67.5°,\ 157.5°$

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Q5. Solve $3\cos^2\theta + 5\sin\theta - 1 = 0$ for $0° \le \theta \le 360°$ . Give answers to 1 decimal place.

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Replace $\cos^2\theta$ using $\cos^2\theta = 1 - \sin^2\theta$ :

3(1-\sin^2\theta) + 5\sin\theta - 1 = 0

3 - 3\sin^2\theta + 5\sin\theta - 1 = 0

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

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

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

Factor 1: $\sin\theta = -\dfrac{1}{3}$ , so $\alpha = \sin^{-1}(0.3333…) = 19.47°$ Solutions in Q3 and Q4:

\theta = 180° + 19.47° = 199.5°, \qquad \theta = 360° - 19.47° = 340.5°

Factor 2: $\sin\theta = 2$ — no solution (since $-1 \le \sin\theta \le 1$ ).

Answer: $\theta = 199.5°,\ 340.5°$

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Connections

Prerequisite subtopics (assumed known):

Exact Trigonometric Values — values of $\sin$ , $\cos$ , $\tan$ at $0°, 30°, 45°, 60°, 90°$ are essential for giving exact answers and recognising principal values without a calculator.
Trigonometric Identities — the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ and the identity $\tan\theta = \sin\theta/\cos\theta$ are frequently needed to reduce equations to a single ratio before solving.

Likely next subtopics:

Further Trigonometric Identities (A-Level Pure Mathematics 3) — double angle formulae such as $\sin 2\theta = 2\sin\theta\cos\theta$ create equations that are solved with exactly this method.
Differentiation of Trigonometric Functions — stationary points of $\sin$ and $\cos$ functions require setting a trigonometric derivative to zero and solving the resulting equation.
Integration of Trigonometric Functions — definite integrals over trigonometric curves require finding where curves intersect axes, again using these solving techniques.

Introduction

Core Concept

The CAST Diagram and Symmetry

Symmetry Rules

Equations with a Transformed Angle

Key Formulae & Definitions

Worked Examples

Example 1 — Basic equation with a double angle

Example 2 — Quadratic in $\cos\theta$ using an identity

Example 3 — Mixed equation reduced using $\tan$

Common Mistakes & Examiner Pitfalls

Practice Questions

Connections

Figures

Keep learning

Introduction

Core Concept

The CAST Diagram and Symmetry

Symmetry Rules

Equations with a Transformed Angle

Key Formulae & Definitions

Worked Examples

Example 1 — Basic equation with a double angle

Example 2 — Quadratic in cos⁡θ\cos\thetacosθ using an identity

Example 3 — Mixed equation reduced using tan⁡\tantan

Common Mistakes & Examiner Pitfalls

Practice Questions

Connections

Figures

Keep learning

Example 2 — Quadratic in $\cos\theta$ using an identity

Example 3 — Mixed equation reduced using $\tan$