(a)[3]
Express $\sqrt{6}\cos\theta + 3\sin\theta$ in the form $R\cos(\theta - \alpha)$, where $R > 0$ and $0^\circ < \alpha < 90^\circ$. Give the exact value of $R$ and quote $\alpha$ correct to 2 decimal places.
(b)[4]
Hence solve the equation $\sqrt{6}\cos\!\left(\tfrac{1}{3}x\right) + 3\sin\!\left(\tfrac{1}{3}x\right) = 2.5$, for $0^\circ < x < 360^\circ$.