(i)[3]
Rewrite $(\sqrt{6})\sin x + \cos x$ in the form $R\sin(x + \alpha)$, where $R > 0$ and $0^{\circ} < \alpha < 90^{\circ}$. Give the exact value of $R$ and state $\alpha$ correct to $3$ decimal places.
(ii)[4]
Hence solve the equation $(\sqrt{6})\sin 2\theta + \cos 2\theta = 2$, for $0^{\circ} < \theta < 180^{\circ}$.