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10.5PureStretch

Harder Trigonometric Equations

Harder equations add a twist: a multiple angle like 2θ2\theta, or a quadratic in sin⁡θ\sin\theta. The methods are the same once you handle the twist carefully — especially the trap of a changed solution range.

35 min Trigonometric Identities and Equations
Edexcel AS Level Maths: 10.5 Harder Trigonometric EquationsCurated by Kepler, not produced by us. Watch it before the notes below.
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What you'll be able to do

  • Solve equations with a multiple angle (e.g. 2θ)
  • Adjust the solution range for multiple angles
  • Solve quadratic trigonometric equations
  • Use a substitution to simplify
1

Multiple angles

For an equation in 2θ2\theta over 0∘≤θ≤360∘0^\circ \le \theta \le 360^\circ, the angle 2θ2\theta runs over 0∘0^\circ–720∘720^\circ. Solve for 2θ2\theta across the wider\textbf{wider} range first, then divide every solution by 22.

Example 1Solve sin⁡2θ=0.5\sin 2\theta = 0.5 for 0∘≤θ≤360∘0^\circ \le \theta \le 360^\circ.
1Let the angle be 2θ2\theta, range 0∘0^\circ–720∘720^\circ.
22θ=30∘,150∘,390∘,510∘2\theta = 30^\circ, 150^\circ, 390^\circ, 510^\circ.
3Divide by 2.
Answerθ=15∘,75∘,195∘,255∘\theta = 15^\circ, 75^\circ, 195^\circ, 255^\circ

Tip — Widen the range to match the multiple angle BEFORE solving, or you will miss solutions.

2

Quadratic trig equations

An equation like 2sin⁡2θ−sin⁡θ−1=02\sin^2\theta - \sin\theta - 1 = 0 is a quadratic in sin⁡θ\sin\theta. Substitute s=sin⁡θs = \sin\theta, solve the quadratic, then solve each resulting simple equation.

Example 1Solve 2sin⁡2θ−sin⁡θ−1=02\sin^2\theta - \sin\theta - 1 = 0 for 0∘≤θ≤360∘0^\circ \le \theta \le 360^\circ.
1Let s=sin⁡θs = \sin\theta: 2s2−s−1=(2s+1)(s−1)=02s^2 - s - 1 = (2s + 1)(s - 1) = 0.
2So sin⁡θ=1\sin\theta = 1 or sin⁡θ=−12\sin\theta = -\tfrac{1}{2}.
3sin⁡θ=1⇒90∘\sin\theta = 1 \Rightarrow 90^\circ; sin⁡θ=−12⇒210∘,330∘\sin\theta = -\tfrac{1}{2} \Rightarrow 210^\circ, 330^\circ.
Answerθ=90∘,210∘,330∘\theta = 90^\circ, 210^\circ, 330^\circ
3

Using an identity first

Sometimes you must replace sin⁡2θ\sin^2\theta with 1−cos⁡2θ1 - \cos^2\theta (or vice versa) so the whole equation is in one ratio, turning it into a solvable quadratic.

Tip — If an equation mixes sin² and cos, use the Pythagorean identity to get everything in terms of one ratio.

Formula recap

for nθ: widen the range by×n\text{for } n\theta: \text{ widen the range by} \times n
Multiple angle range.
2s2−s−1=0, s=sin⁡θ2s^{2} - s - 1 = 0,\ s = \sin\theta
Quadratic in a ratio.
sin⁡2θ=1−cos⁡2θ\sin^{2}\theta = 1 - \cos^{2}\theta
Identity to unify ratios.

Common mistakes to avoid

Solving sin 2θ = 0.5 in 0°–360° and dividing — missing half the solutions.
Solve over 0°–720° for 2θ first, then divide each by 2.
Discarding sin θ = 1 because it “only gives one solution”.
That is fine — sin θ = 1 gives θ = 90° in range; include it.

Key takeaways

  • Multiple angle nθ: widen the solution range ×n, solve, then divide.
  • Quadratic trig equations: substitute s = sin θ (or cos θ), solve, then unwind.
  • Use the Pythagorean identity to get a single ratio when needed.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Check your understanding

Work through questions on Harder Trigonometric Equations. Use the feedback to decide what needs another look.

7 past-paper questions · 24 marks

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