(a)[3]
Rewrite $3\sin\theta + 2\cos\theta$ as $R\sin(\theta + \alpha)$, with $R > 0$ and $0^\circ < \alpha < 90^\circ$; give the exact value of $R$ and the value of $\alpha$ correct to 2 decimal places.
(b)[3]
Hence solve the equation $3\sin\theta + 2\cos\theta = 1$ for $0^\circ < \theta < 180^\circ$.