(a) To express \(5 \sin \theta + 12 \cos \theta\) in the form \(R \cos(\theta - \alpha)\), we use the identity:

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

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

Here, \(a = 5\) and \(b = 12\).

Calculate \(R\):

\(R = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{12}{5}\)

\(\alpha = \tan^{-1}\left(\frac{12}{5}\right) \approx 0.395\)

Thus, \(5 \sin \theta + 12 \cos \theta = 13 \cos(\theta - 0.395)\).

(b) Solve \(5 \sin 2x + 12 \cos 2x = 6\) using the result from part (a):

\(5 \sin 2x + 12 \cos 2x = 13 \cos(2x - 0.395)\)

Set \(13 \cos(2x - 0.395) = 6\):

\(\cos(2x - 0.395) = \frac{6}{13}\)

\(2x - 0.395 = \cos^{-1}\left(\frac{6}{13}\right)\)

\(2x = \cos^{-1}\left(\frac{6}{13}\right) + 0.395\)

Calculate \(\cos^{-1}\left(\frac{6}{13}\right)\):

\(\cos^{-1}\left(\frac{6}{13}\right) \approx 1.091\)

\(2x = 1.091 + 0.395\)

\(2x = 1.486\)

\(x = \frac{1.486}{2} \approx 0.743\)

For the second solution:

\(2x = 2\pi - 1.091 + 0.395\)

\(2x = 5.587\)

\(x = \frac{5.587}{2} \approx 2.79\)

Thus, the solutions are \(x = 0.743\) or \(x = 2.79\).