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