Trig identities and equations

The idea

Two identities turn many equations into ones you can solve: $\tan x \equiv \frac{\sin x}{\cos x}$ and $\sin^2 x + \cos^2 x \equiv 1$ . Use them to get one trig function, then solve like an ordinary equation.

Replace $\cos^2 x$ with $1 - \sin^2 x$ (or the other way) so only one function is left.
Rearrange to $= 0$ and factorise.
Solve each factor in the interval.

Watch out Never divide by $\sin x$ or $\cos x$ : factorise instead, or you lose solutions.

See it worked

Worked example

Solve $2\cos^2 x + \sin x = 1$ for $0^\circ \le x < 360^\circ$ .

$2(1 - \sin^2 x) + \sin x - 1 = 0$ , so $2\sin^2 x - \sin x - 1 = 0$ .
$(2\sin x + 1)(\sin x - 1) = 0$ .
$\sin x = 1$ : $x = 90^\circ$ . $\sin x = -\frac{1}{2}$ : $x = 210^\circ, 330^\circ$ .

Answer $x = 90^\circ, 210^\circ, 330^\circ$

Your turn

Question 1Non-calculator3 marks

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

Question 2Calculator allowed3 marks

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

Question 3Non-calculator3 marks

$\sin x = \frac{3}{5}$ and $x$ is obtuse. Find the exact value of $\cos x$ .

In past papers

32 official questions in 30 papers since 2018. The latest:

All A-level Maths past papers

A-level Maths Pro, Pearson Edexcel 9MA0

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