If \({1\over 6}sin\theta\), \(cos\theta\) and \(tan\theta\) are in G.P. then the general solution for \(\theta\) is

Solution :

Since, \({1\over 6}sin\theta\), \(cos\theta\) and \(tan\theta\) are in G.P.

\(\implies\) \(cos^2\theta\) = \({1\over 6}sin\theta\).\(cos\theta\)

\(\implies\) \(6cos^3\theta\) + \(cos^2\theta\) – 1 = 0

\(\therefore\) (\(2cos\theta – 1\))(\(3cos^2\theta\) + \(2cos\theta\) + 1) = 0

\(cos\theta\) = \(1\over 2\) (other values are imaginary)

\(cos\theta\) = \(cos\pi\over 3\)

\(\theta\) = \(2n\pi \pm {\pi\over 3}\), n \(\in\) I