(i) To express \(\sqrt{6} \sin x + \cos x\) in the form \(R \sin(x + \alpha)\), we use the identity:

\(a \sin x + b \cos x \equiv R \sin(x + \alpha)\)

where \(R = \sqrt{a^2 + b^2}\) and \(\tan \alpha = \frac{b}{a}\).

Here, \(a = \sqrt{6}\) and \(b = 1\).

Calculate \(R\):

\(R = \sqrt{(\sqrt{6})^2 + 1^2} = \sqrt{6 + 1} = \sqrt{7}\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{1}{\sqrt{6}}\)

\(\alpha = \tan^{-1}\left(\frac{1}{\sqrt{6}}\right) \approx 22.208^\circ\)

(ii) Solve \(\sqrt{6} \sin 2\theta + \cos 2\theta = 2\).

Express in the form \(R \sin(2\theta + \alpha)\):

\(\sqrt{6} \sin 2\theta + \cos 2\theta = \sqrt{7} \sin(2\theta + 22.208^\circ)\)

Set equal to 2:

\(\sqrt{7} \sin(2\theta + 22.208^\circ) = 2\)

\(\sin(2\theta + 22.208^\circ) = \frac{2}{\sqrt{7}}\)

Calculate \(2\theta + 22.208^\circ\):

\(2\theta + 22.208^\circ = \sin^{-1}\left(\frac{2}{\sqrt{7}}\right) \approx 49.107^\circ\)

\(2\theta = 49.107^\circ - 22.208^\circ\)

\(2\theta = 26.899^\circ\)

\(\theta = 13.4495^\circ \approx 13.4^\circ\)

For the second solution:

\(2\theta + 22.208^\circ = 180^\circ - 49.107^\circ\)

\(2\theta = 180^\circ - 49.107^\circ - 22.208^\circ\)

\(2\theta = 108.685^\circ\)

\(\theta = 54.3425^\circ \approx 54.3^\circ\)