Solve \(\sqrt 3 \cos x + \sin x = \sqrt 2\), for \(0 \le x \le 2\pi\).

First of all we need to put \(\sqrt 3 \cos x + \sin x\) into the form \(k\cos (x - \alpha)\).

\(k\cos (x-\alpha) = k\cos x\cos \alpha + k\sin x\sin \alpha\)

\(k\cos \alpha = \sqrt 3\)

\(k\sin \alpha = 1\)

Calculate \(k\) using the values for the coefficients:

\(k = \sqrt {\left( {\sqrt {{3^2}} } \right) + {1^2}}\)

\(= \sqrt {3 + 1}\)

\(= \sqrt 4\)

\(= 2\)

Since \(\tan \theta = \frac{{sin\theta }}{{\cos \theta }}\)

\(\tan \alpha = \frac{1}{{\sqrt 3 }}\)

\(\alpha = \frac{\pi }{6}\)

Therefore \(\sqrt 3 \cos x + \sin x = 2\cos \left( {x - \frac{\pi }{6}} \right)\)

\(2\cos \left( {x - \frac{\pi }{6}} \right) = \sqrt 2\)

\(\cos \left( {x - \frac{\pi }{6}} \right) = \frac{{\sqrt 2 }}{2}\)

\(\cos \left( {x - \frac{\pi }{6}} \right) = \frac{1}{{\sqrt 2 }}\)

\(x - \frac{\pi }{6} = {\cos ^{ - 1}}\frac{1}{{\sqrt 2 }}\)

\(x - \frac{\pi }{6} = \frac{\pi }{4}\)

\(x = \frac{{5\pi }}{{12}}\)

\(x - \frac{\pi }{6} = 2\pi - \frac{\pi }{4}\)

\(x - \frac{\pi }{6} = \frac{{7\pi }}{4}\)

\(x = \frac{{23\pi }}{{12}}\)

Therefore \(x = \frac{{5\pi }}{{12}},\,\frac{{23\pi }}{{12}}\)