(a)[3]
Express $\sqrt{2} \cos x - \sqrt{5} \sin x$ as $R \cos (x + \alpha)$, where $R > 0$ and $0^{\circ} < \alpha < 90^{\circ}$. State the exact value of $R$ and the value of $\alpha$ accurate to 3 d.p.
(b)[4]
Hence solve the equation $\sqrt{2} \cos 2\theta - \sqrt{5} \sin 2\theta = 1$, for $0^{\circ} < \theta < 180^{\circ}$.