(i)[3]
Rewrite $(\sqrt{6})\cos \theta + (\sqrt{10})\sin \theta$ in the form $R\cos(\theta - \alpha)$, where $R > 0$ and $0^\circ < \alpha < 90^\circ$. State the value of $\alpha$ correct to 2 decimal places.
(ii(a))[2]
Hence, determine the least positive angle $\theta$ satisfying $(\sqrt{6})\cos \theta + (\sqrt{10})\sin \theta = -4$.
(ii(b))[4]
Hence, determine the least positive angle $\theta$ satisfying $(\sqrt{6})\cos \frac{1}{2}\theta + (\sqrt{10})\sin \frac{1}{2}\theta = 3$.