(i) Express \(\sqrt{6} \sin x + \cos x\) in the form \(R \sin(x + \alpha)\), where \(R > 0\) and \(0^\circ < \alpha < 90^\circ\). State the exact value of \(R\) and give \(\alpha\) correct to 3 decimal places.

(ii) Hence solve the equation \(\sqrt{6} \sin 2\theta + \cos 2\theta = 2\), for \(0^\circ < \theta < 180^\circ\).