(i) To express \(5 \sin x + 12 \cos x\) 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, \(a = 5\) and \(b = 12\).

Calculate \(R\):

\(R = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{12}{5}\)

\(\alpha = \tan^{-1}\left(\frac{12}{5}\right) \approx 67.38^\circ\)

(ii) Solve \(5 \sin 2\theta + 12 \cos 2\theta = 11\).

Using the result from part (i), express \(5 \sin 2\theta + 12 \cos 2\theta\) as:

\(13 \sin(2\theta + 67.38^\circ) = 11\)

\(\sin(2\theta + 67.38^\circ) = \frac{11}{13}\)

Find \(2\theta + 67.38^\circ\):

\(2\theta + 67.38^\circ = \sin^{-1}\left(\frac{11}{13}\right) \approx 57.80^\circ\)

\(2\theta + 67.38^\circ = 180^\circ - 57.80^\circ = 122.20^\circ\)

Calculate \(2\theta\):

\(2\theta = 57.80^\circ - 67.38^\circ = -9.58^\circ\)

\(2\theta = 122.20^\circ - 67.38^\circ = 54.82^\circ\)

Since \(2\theta\) must be positive, use \(54.82^\circ\):

\(\theta = \frac{54.82^\circ}{2} = 27.41^\circ\)

Find the second solution:

\(2\theta = 180^\circ + 54.82^\circ = 234.82^\circ\)

\(\theta = \frac{234.82^\circ}{2} = 117.41^\circ\)

Thus, the solutions are \(\theta \approx 27.4^\circ, 117.4^\circ\).