Solve the equation $\sin 2x - \cos 2x = 1 + \sin x - \cos x$ for $x \in [ - \pi ,{\text{ }}\pi ]$.

$(\sin 2x - \sin x) - (\cos 2x - \cos x) = 1$

attempt to use both double-angle formulae, in whatever form     M1

$(2\sin x\cos x - \sin x) - (2{\cos ^2}x - 1 - \cos x) = 1$

or $(2\sin x\cos x - \sin x) - (2{\cos ^2}x - \cos x) = 0$ for example     A1

Note:     Allow any rearrangement of the above equations.

$\sin x(2\cos x - 1) - \cos x(2\cos x - 1) = 0$

$(\sin x - \cos x)(2\cos x - 1) = 0$     (M1)

$\tan x = 1{\text{ and }}\cos x = \frac{1}{2}$     A1A1

Note:     These A marks are dependent on the M mark awarded for factorisation.

$x =  - \frac{{3\pi }}{4},{\text{ }} - \frac{\pi }{3},{\text{ }}\frac{\pi }{3},{\text{ }}\frac{\pi }{4}$     A2

Note:     Award A1 for two correct answers, which could be for both tan or both cos solutions, for example.

[7 marks]