(i)[3]
Express $(\sqrt{5})\cos\theta - 2\sin\theta$ in the form $R\cos(\theta + \alpha)$, with $R > 0$ and $0^\circ < \alpha < 90^\circ$. State the value of $\alpha$ correct to $2$ decimal places.
(ii(a))[4]
Hence solve the equation $(\sqrt{5})\cos\theta - 2\sin\theta = 0.9$ for $0^\circ < \theta < 360^\circ$.
(ii(b))[2]
State the maximum and minimum values of $10 + (\sqrt{5})\cos\theta - 2\sin\theta$ as $\theta$ changes.