(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 \(0^\circ < \theta < 180^\circ\)).