(i) To express \(\cos x + 3 \sin x\) in the form \(R \cos(x - \alpha)\), we use the identity:

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

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

Here, \(a = 1\) and \(b = 3\).

Calculate \(R\):

\(R = \sqrt{1^2 + 3^2} = \sqrt{10}\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{3}{1} = 3\)

\(\alpha = \tan^{-1}(3) \approx 71.57^\circ\)

(ii) Solve \(\cos 2\theta + 3 \sin 2\theta = 2\).

Using the result from part (i), express \(\cos 2\theta + 3 \sin 2\theta\) as:

\(\sqrt{10} \cos(2\theta - 71.57^\circ) = 2\)

\(\cos(2\theta - 71.57^\circ) = \frac{2}{\sqrt{10}}\)

\(2\theta - 71.57^\circ = \cos^{-1}\left(\frac{2}{\sqrt{10}}\right)\)

\(2\theta - 71.57^\circ \approx 50.77^\circ\)

\(2\theta \approx 50.77^\circ + 71.57^\circ = 122.34^\circ\)

\(\theta \approx 61.17^\circ\)

For the second solution:

\(2\theta \approx 360^\circ - 50.77^\circ + 71.57^\circ = 380.8^\circ\)

\(\theta \approx 190.4^\circ\)

Since \(\theta\) must be between \(0^\circ\) and \(90^\circ\), the valid solutions are:

\(\theta = 10.4^\circ, 61.2^\circ\)