Solution :
We have,
\(\sqrt{3} cos \theta\) + \(sin \theta\) = \(\sqrt{2}\) ………….(i)
This is of the form \(a cos\theta\) + \(b sin \theta\) = c, where a = \(\sqrt{3}\), b = 1 and c = \(\sqrt{2}\).
Let a = \(r cos\alpha\) and b = \(r sin\alpha\). Then,
\(\sqrt{3}\) = \(r cos\alpha\) and 1 = \(r sin\alpha\)
\(\implies\) r = \(\sqrt{a^2 + b^2}\) = \(\sqrt{(\sqrt{3})^2 + 1^2}\) = 2 and \(tan \alpha\) = \(1\over \sqrt{3}\) \(\implies\) \(\alpha\) = \(\pi\over 6\)
Substituting a = \(\sqrt{3}\) = \(r cos\alpha\) and b = 1 = \(r sin \alpha\) in the equation (i) it reduces to
\(r cos\alpha cos\theta\) + \(r sin\alpha sin\theta\) = \(\sqrt{2}\)
\(\implies\) \(r cos(\theta – \alpha)\) = \(\sqrt{2}\)
\(\implies\) \(2 cos(\theta – {\pi\over 6}\) = \(\sqrt{2}\)
\(\implies\) \(cos(\theta – {\pi\over 6})\) = \(1\over \sqrt{2}\)
\(\implies\) \(cos(\theta – {\pi\over 6})\) = \(cos{\pi\over 4}\)
\(\implies\) \(\theta – {\pi\over 6}\) = \(2n\pi \pm {\pi\over 4}\), n \(\in\) Z.
\(\implies\) \(\theta\) = \(2n\pi \pm {\pi\over 4} + {\pi\over 6}\), n \(\in\) Z.
\(\implies\) \(\theta\) = \(2n\pi + {\pi\over 4} + {\pi\over 6}\) or, \(\theta\) = \(2n\pi – {\pi\over 4} + {\pi\over 6}\)
\(\implies\) \(\theta\) = \(2n\pi + {5\pi\over 12}\) or \(2n\pi – {\pi\over 12}\)
Hence, \(\theta\) = \(2n\pi + {5\pi\over 12}\) or, \(\theta\) = \(2n\pi – {5\pi\over 12}\), where n \(\in\) Z.
