(i)[3]
Express $2 \cos \theta + \sqrt{5} \sin \theta$ in the form $R \cos(\theta - \alpha)$ where $R > 0$ and $0^\circ < \alpha < 90^\circ$, and give the value of $\alpha$ correct to 2 decimal places.
(ii)[4]
Hence determine the solutions of the equation $2 \cos \theta + \sqrt{5} \sin \theta = 1$ for $0^\circ < \theta < 360^\circ$.