(a) To express \(4 \cos x - \sin x\) in the form \(R \cos(x + \alpha)\), 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, \(a = 4\) and \(b = 1\).

Calculate \(R\):

\(R = \sqrt{4^2 + 1^2} = \sqrt{16 + 1} = \sqrt{17}\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{1}{4}\)

\(\alpha = \tan^{-1}\left(\frac{1}{4}\right) \approx 14.04^\circ\)

(b) Solve \(4 \cos 2x - \sin 2x = 3\) using the result from part (a):

\(R \cos(2x + \alpha) = 3\)

\(\sqrt{17} \cos(2x + 14.04^\circ) = 3\)

\(\cos(2x + 14.04^\circ) = \frac{3}{\sqrt{17}}\)

Calculate \(2x + 14.04^\circ\):

\(2x + 14.04^\circ = \cos^{-1}\left(\frac{3}{\sqrt{17}}\right)\)

\(2x + 14.04^\circ \approx 43.31^\circ\)

\(2x \approx 43.31^\circ - 14.04^\circ = 29.27^\circ\)

\(x \approx 14.6^\circ\)

For the second solution:

\(2x + 14.04^\circ = 360^\circ - 43.31^\circ\)

\(2x + 14.04^\circ \approx 316.69^\circ\)

\(2x \approx 316.69^\circ - 14.04^\circ = 302.65^\circ\)

\(x \approx 151.3^\circ\)