(a) To express \(3 \cos x + 2 \cos(x - 60^\circ)\) in the form \(R \cos(x - \alpha)\):

First, expand \(2 \cos(x - 60^\circ)\):

\(2 \cos(x - 60^\circ) = 2(\cos x \cos 60^\circ + \sin x \sin 60^\circ) = \cos x + \sqrt{3} \sin x\)

Combine with \(3 \cos x\):

\(3 \cos x + \cos x + \sqrt{3} \sin x = 4 \cos x + \sqrt{3} \sin x\)

Now, 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, \(a = 4\) and \(b = \sqrt{3}\):

\(R = \sqrt{4^2 + (\sqrt{3})^2} = \sqrt{16 + 3} = \sqrt{19}\) \(\tan \alpha = \frac{\sqrt{3}}{4}\)

Thus, \(\alpha = \tan^{-1}\left(\frac{\sqrt{3}}{4}\right) \approx 23.41^\circ\).

(b) Solve \(3 \cos 2\theta + 2 \cos(2\theta - 60^\circ) = 2.5\):

Using the result from part (a), express:

\(3 \cos 2\theta + 2 \cos(2\theta - 60^\circ) = \sqrt{19} \cos(2\theta - 23.41^\circ)\)

Set equal to 2.5:

\(\sqrt{19} \cos(2\theta - 23.41^\circ) = 2.5\)

\(\cos(2\theta - 23.41^\circ) = \frac{2.5}{\sqrt{19}}\)

Find \(2\theta\):

\(2\theta - 23.41^\circ = \cos^{-1}\left(\frac{2.5}{\sqrt{19}}\right)\)

Calculate \(2\theta\):

\(2\theta = \cos^{-1}\left(\frac{2.5}{\sqrt{19}}\right) + 23.41^\circ\)

\(\theta = 39.2^\circ\) and \(\theta = 82.1^\circ\) (considering the range \(0^\circ < \theta < 180^\circ\)).

(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:

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

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

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

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

(i) Start with the equation:

\(\sqrt{2} \csc x + \cot x = \sqrt{3}\)

Express \(\csc x\) and \(\cot x\) in terms of sine and cosine:

\(\sqrt{2} \left( \frac{1}{\sin x} \right) + \frac{\cos x}{\sin x} = \sqrt{3}\)

Combine into a single fraction:

\(\frac{\sqrt{2} + \cos x}{\sin x} = \sqrt{3}\)

Multiply through by \(\sin x\):

\(\sqrt{2} + \cos x = \sqrt{3} \sin x\)

Rearrange to:

\(\sqrt{3} \sin x - \cos x = \sqrt{2}\)

Use the identity:

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

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

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

Calculate \(R\):

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

Calculate \(\alpha\):

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

\(\alpha = 30^\circ\)

Thus, \(\sqrt{3} \sin x - \cos x = 2 \sin(x - 30^\circ)\).

(ii) Solve \(2 \sin(x - 30^\circ) = \sqrt{2}\):

\(\sin(x - 30^\circ) = \frac{\sqrt{2}}{2}\)

\(x - 30^\circ = 45^\circ\) or \(x - 30^\circ = 135^\circ\)

\(x = 75^\circ\) or \(x = 165^\circ\)

(i) Expand \(2 \sin(x - 30^\circ)\) using the angle subtraction formula:

\(2 \sin(x - 30^\circ) = 2(\sin x \cos 30^\circ - \cos x \sin 30^\circ)\)

\(= \sqrt{3} \sin x - \cos x\)

Combine terms: \(\sqrt{3} \sin x - 2 \cos x\).

Using the identity:

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

\(R = \sqrt{a^2 + b^2} = \sqrt{(\sqrt{3})^2 + (-2)^2} = \sqrt{7}\)

\(\tan \alpha = \frac{b}{a} = \frac{-2}{\sqrt{3}}\)

\(\alpha = \tan^{-1}\left(\frac{-2}{\sqrt{3}}\right) \approx 49.11^\circ\)

(ii) Solve \(\sqrt{3} \sin x - 2 \cos x = 1\):

\(R \sin(x - \alpha) = 1\)

\(\sqrt{7} \sin(x - 49.11^\circ) = 1\)

\(\sin(x - 49.11^\circ) = \frac{1}{\sqrt{7}}\)

\(x - 49.11^\circ = \sin^{-1}\left(\frac{1}{\sqrt{7}}\right) \approx 22.21^\circ\)

\(x = 22.21^\circ + 49.11^\circ = 71.3^\circ\)

(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 \(x = 110.8^\circ\).