(i)[3]
Express $4\sin\theta - 6\cos\theta$ as $R\sin(\theta - \alpha)$, with $R > 0$ and $0^\circ < \alpha < 90^\circ$. Give the exact value of $R$ and the value of $\alpha$ accurate to 2 decimal places.
(ii)[4]
Solve $4\sin\theta - 6\cos\theta = 3$ for $0^\circ \leq \theta \leq 360^\circ$.
(iii)[2]
Find the maximum and minimum possible values of $(4\sin\theta - 6\cos\theta)^2 + 8$ as $\theta$ varies.