Solving Basic Trigonometric Equations Revision Notes | GCSE AQA Higher Further Maths

1. Why Trig Equations Have More Than One Answer

You already know how to find a missing angle in a right-angled triangle using $\sin$, $\cos$, or $\tan$. But those triangle problems always give one tidy answer. Trigonometric equations are different — within a full $360^\circ$ interval, there are usually two solutions, and missing one of them costs marks.

The reason comes directly from the shape of the trig graphs. The sine curve, for example, reaches the same height twice in every $360^\circ$ cycle — once on the way up and once on the way down. So if $\sin x = 0.5$, there are two angles between $0^\circ$ and $360^\circ$ that work. Your job is to find both.

2. The CAST Diagram and Quadrant Rules

The CAST diagram divides the full $360^\circ$ rotation into four quadrants and tells you where each ratio is positive. Reading anticlockwise from the bottom-right: Cosine is positive in Q4 ($270^\circ$–$360^\circ$), All are positive in Q1 ($0^\circ$–$90^\circ$), Sine is positive in Q2 ($90^\circ$–$180^\circ$), Tangent is positive in Q3 ($180^\circ$–$270^\circ$).

Once you know which two quadrants hold your solutions, you use the reference angle — the acute angle from your calculator — to build the actual solutions. The symmetry rules for each ratio are fixed and worth memorising.

3. Solving a Sine Equation — Worked Example

Theory clicks into place when you see the method in action. The process is always the same four moves: find the reference angle, identify the correct quadrants using CAST, apply the symmetry rule, and check both answers sit inside the given interval.

Worked example: Solving a Sine Equation

Solve $\sin x = 0.5$ for $0^\circ \leq x \leq 360^\circ$.

1. $\alpha = \sin^{-1}(0.5) = 30^\circ$ — Always use the positive value of $k$ in the inverse function to get the acute reference angle. [M1]
2. $k = 0.5 > 0$, so sine is positive. Sine is positive in Q1 and Q2. — From the CAST diagram: S is in Q2, A (all) covers Q1. Both quadrants give positive sine.
3. $x = \alpha = 30^\circ$ — Q1 solution is always the reference angle itself.
4. $x = 180^\circ - \alpha = 180^\circ - 30^\circ = 150^\circ$ — Q2 solution for sine: subtract the reference angle from $180^\circ$. [A1]
5. $30^\circ$ and $150^\circ$ both satisfy $0^\circ \leq x \leq 360^\circ$. $\checkmark$ — Both solutions are valid. Always verify before writing your final answer.

Answer: $x = 30^\circ$ or $x = 150^\circ$

The graph confirms it: the horizontal line $y = 0.5$ cuts the sine curve at exactly two points between $0^\circ$ and $360^\circ$ — at $x = 30^\circ$ and $x = 150^\circ$. This visual check is a powerful way to sense-check your algebra.

4. Solving a Cosine Equation — Worked Example

Cosine has its own symmetry pattern. Where sine is symmetric about $90^\circ$, cosine is symmetric about $0^\circ$ (and $360^\circ$). That gives the second solution as $360^\circ - \alpha$ rather than $180^\circ - \alpha$.

Worked example: Solving a Cosine Equation

Solve $\cos x = -0.6$ for $0^\circ \leq x \leq 360^\circ$. Give your answers to 1 decimal place.

1. $\alpha = \cos^{-1}(0.6) = 53.1^\circ$ (1 d.p.) — Use the positive value $|{-0.6}| = 0.6$ in the inverse cosine. [M1]
2. $k = -0.6 < 0$, so cosine is negative. Cosine is negative in Q2 and Q3. — Cosine is positive in Q1 and Q4, so it is negative in the other two quadrants: Q2 and Q3.
3. $x = 180^\circ - \alpha = 180^\circ - 53.1^\circ = 126.9^\circ$ — Q2 solution for negative cosine: $180^\circ - \alpha$. [A1]
4. $x = 180^\circ + \alpha = 180^\circ + 53.1^\circ = 233.1^\circ$ — Q3 solution for negative cosine: $180^\circ + \alpha$. [A1]
5. $126.9^\circ$ and $233.1^\circ$ both satisfy $0^\circ \leq x \leq 360^\circ$. $\checkmark$ — Both solutions are valid and rounded to 1 decimal place as required.

Answer: $x = 126.9^\circ$ or $x = 233.1^\circ$

5. Solving a Tangent Equation — Worked Example

Tangent has period 180°, so its solutions in 0° to 360° always come in pairs that are 180° apart. For positive tangent, the solutions lie in Q1 and Q3, so they are x = α and x = α + 180°. For negative tangent, the solutions lie in Q2 and Q4, so they are x = 180° - α and x = 360° - α.

Worked example: Solving a Tangent Equation

Solve $\tan x = \sqrt{3}$ for $0^\circ \leq x \leq 360^\circ$.

1. $\alpha = \tan^{-1}(\sqrt{3}) = 60^\circ$ — $\tan 60^\circ = \sqrt{3}$ is an exact value worth knowing. [M1]
2. $k = \sqrt{3} > 0$, so tangent is positive. Tangent is positive in Q1 and Q3. — From CAST: T is in Q3, A (all) covers Q1.
3. $x = \alpha = 60^\circ$ — Q1 solution is the reference angle. [A1]
4. $x = 180^\circ + \alpha = 180^\circ + 60^\circ = 240^\circ$ — Q3 solution for positive tangent: add $180^\circ$ to the reference angle. [A1]
5. $60^\circ$ and $240^\circ$ both satisfy $0^\circ \leq x \leq 360^\circ$. $\checkmark$ — Both solutions confirmed. Note they are exactly $180^\circ$ apart — this is always true for tangent.

Answer: $x = 60^\circ$ or $x = 240^\circ$

6. Practice — Try These Yourself

You have now seen the method applied to all three ratios. The structure is identical each time: reference angle, CAST quadrants, symmetry rule, interval check. Work through these problems before looking at the answers.

7. Exam Technique and Common Mistakes

Examiners report that the most common source of lost marks on these questions is not the algebra — it is forgetting the second solution entirely. A student who writes only $x = 30^\circ$ for $\sin x = 0.5$ scores one mark out of two. The method is correct; the answer is incomplete.

A second trap appears when the equation has a negative right-hand side. Many students apply the positive-case symmetry rule and land in the wrong quadrants. Always check the sign of $k$ before choosing your quadrants — the CAST diagram is your safeguard.