Solve : \(\sqrt{3} cos \theta\) + \(sin \theta\) = \(\sqrt{2}\)

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.

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