(a)[5]
Express $3\sqrt{3}\sin\left(\theta + \frac{\pi}{6}\right) - 2\sin\theta$ in the form $R\sin(\theta + \alpha)$, where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$. State the exact value of $R$ and give the value of $\alpha$ correct to 3 decimal places.
(b)[4]
Hence, solve $3\sqrt{3}\sin\left(2x+\frac{\pi}{6}\right)-2\sin 2x=\sqrt{6}$ for $0<x<\pi$.