(i) To express \(8 \cos \theta - 15 \sin \theta\) in the form \(R \cos(\theta + \alpha)\), we use the identity:

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

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

Here, \(a = 8\) and \(b = -15\).

Calculate \(R\):

\(R = \sqrt{8^2 + (-15)^2} = \sqrt{64 + 225} = \sqrt{289} = 17\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{-15}{8}\)

\(\alpha = \tan^{-1}\left(\frac{-15}{8}\right) \approx 61.93^\circ\)

(ii) Solve \(8 \cos 2x - 15 \sin 2x = 4\).

Using the result from part (i), express the equation as:

\(17 \cos(2x + 61.93^\circ) = 4\)

\(\cos(2x + 61.93^\circ) = \frac{4}{17}\)

\(2x + 61.93^\circ = \cos^{-1}\left(\frac{4}{17}\right) \approx 76.39^\circ\)

\(2x = 76.39^\circ - 61.93^\circ = 14.46^\circ\)

\(x = \frac{14.46^\circ}{2} = 7.23^\circ\)

For the second solution:

\(2x + 61.93^\circ = 360^\circ - 76.39^\circ = 283.61^\circ\)

\(2x = 283.61^\circ - 61.93^\circ = 221.68^\circ\)

\(x = \frac{221.68^\circ}{2} = 110.84^\circ\)

Thus, the solutions are \(x = 7.2^\circ\) and \(x = 110.8^\circ\).