(i)[5]
Start by expanding $2\sin(x - 30^\circ)$, then rewrite $2\sin(x - 30^\circ) - \cos x$ in the form $R\sin(x - \alpha)$, where $R > 0$ and $0^\circ < \alpha < 90^\circ$. State the exact value of $R$ and the value of $\alpha$ correct to 2 decimal places.
(ii)[3]
Hence solve $2\sin(x - 30^\circ) - \cos x = 1$, for $0^\circ < x < 180^\circ$.