Harder trig equations

The idea

Use identities to rewrite an equation in one function of one angle, then factorise. Equations with $2x$ have twice as many solutions in the interval, so solve for $2x$ first over the doubled range.

Replace $\sin 2x$ , $\cos 2x$ or $\sec^2 x$ using an identity.
Factorise; keep every factor, including $\sin x = 0$ or $\cos x = 0$ .
List every solution in the interval.

Watch out Dividing by $\sin x$ or $\cos x$ throws away solutions: factorise instead.

See it worked

Worked example

Solve $2\sin 2\theta = \cos\theta$ for $0^\circ \le \theta < 360^\circ$ .

$4\sin\theta\cos\theta - \cos\theta = 0$ , so $\cos\theta(4\sin\theta - 1) = 0$ .
$\cos\theta = 0$ : $90^\circ, 270^\circ$ .
$\sin\theta = \frac{1}{4}$ : $14.5^\circ, 165.5^\circ$ .

Answer $\theta = 14.5^\circ, 90^\circ, 165.5^\circ, 270^\circ$

Your turn

Question 1Non-calculator3 marks

Solve $\sin 2x = \sin x$ for $0^\circ \le x \le 180^\circ$ .

Question 2Non-calculator3 marks

Solve $\tan 2x = 1$ for $0^\circ \le x < 180^\circ$ .

Question 3Calculator allowed3 marks

Solve $3\cos^2 x + 5\sin x = 1$ for $0^\circ \le x < 360^\circ$ . Give answers to 1 decimal place.

In past papers

40 official questions in 37 papers since 2018. The latest:

All A-level Maths past papers

A-level Maths Pro, Pearson Edexcel 9MA0

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