Solve ${\log _2}(2\sin x) + {\log _2}(\cos x) =  - 1$, for $2\pi  < x < \frac{{5\pi }}{2}$.

correct application of $\log a + \log b = \log ab$     (A1)

eg$\,\,\,\,\,$${\log _2}(2\sin x\cos x),{\text{ }}\log 2 + \log (\sin x) + \log (\cos x)$

correct equation without logs     A1

eg$\,\,\,\,\,$$2\sin x\cos x = {2^{ - 1}},{\text{ }}\sin x\cos x = \frac{1}{4},{\text{ }}\sin 2x = \frac{1}{2}$

recognizing double-angle identity (seen anywhere)     A1

eg$\,\,\,\,\,$$\log (\sin 2x),{\text{ }}2\sin x\cos x = \sin 2x,{\text{ }}\sin 2x = \frac{1}{2}$

evaluating ${\sin ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{6}{\text{ }}(30^\circ )$     (A1)

correct working     A1

eg$\,\,\,\,\,$$x = \frac{\pi }{{12}} + 2\pi ,{\text{ }}2x = \frac{{25\pi }}{6},{\text{ }}\frac{{29\pi }}{6},{\text{ }}750^\circ ,{\text{ }}870^\circ ,{\text{ }}x = \frac{\pi }{{12}}$and $x = \frac{{5\pi }}{{12}}$, one correct final answer

$x = \frac{{25\pi }}{{12}},{\text{ }}\frac{{29\pi }}{{12}}$ (do not accept additional values)     A2     N0

[7 marks]