(a) Expand \(\cos(x - 60^\circ)\) using the formula \(\cos(A - B) = \cos A \cos B + \sin A \sin B\):

\(\cos(x - 60^\circ) = \cos x \cos 60^\circ + \sin x \sin 60^\circ\)

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

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

Substitute into \(2\cos(x - 60^\circ) + \cos x\):

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

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

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

Now, express in the form \(R\cos(x - \alpha)\):

\(R = \sqrt{2^2 + (\sqrt{3})^2} = \sqrt{4 + 3} = \sqrt{7}\)

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

\(\alpha = \tan^{-1}\left(\frac{\sqrt{3}}{2}\right) \approx 40.89^\circ\)

(b) The expression \(R\cos(x - \alpha)\) takes its least value when \(\cos(x - \alpha) = -1\):

\(x - \alpha = 180^\circ\)

\(x = 180^\circ + \alpha\)

\(x = 180^\circ + 40.89^\circ = 220.89^\circ\)