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]