(a) To express \(\sqrt{2} \cos x - \sqrt{5} \sin x\) in the form \(R \cos(x + \alpha)\), we use the identity:

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

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

Here, \(a = \sqrt{2}\) and \(b = -\sqrt{5}\).

Calculate \(R\):

\(R = \sqrt{(\sqrt{2})^2 + (-\sqrt{5})^2} = \sqrt{2 + 5} = \sqrt{7}\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{-\sqrt{5}}{\sqrt{2}}\)

\(\alpha = \tan^{-1}\left(\frac{-\sqrt{5}}{\sqrt{2}}\right) \approx 57.688^\circ\)

(b) Solve \(\sqrt{2} \cos 2\theta - \sqrt{5} \sin 2\theta = 1\).

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

\(\sqrt{7} \cos(2\theta + 57.688^\circ) = 1\)

\(\cos(2\theta + 57.688^\circ) = \frac{1}{\sqrt{7}}\)

Find \(2\theta + 57.688^\circ\):

\(2\theta + 57.688^\circ = \cos^{-1}\left(\frac{1}{\sqrt{7}}\right) \approx 67.792^\circ\)

\(2\theta = 67.792^\circ - 57.688^\circ\)

\(2\theta = 10.104^\circ\)

\(\theta = 5.1^\circ\)

For the second solution:

\(2\theta + 57.688^\circ = 360^\circ - 67.792^\circ\)

\(2\theta = 292.208^\circ - 57.688^\circ\)

\(2\theta = 234.52^\circ\)

\(\theta = 117.3^\circ\)