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


Similar Questions

Find the number of solutions of tanx + secx = 2cosx in [0, \(2\pi\)].

Solve : cos3x + sin2x – sin4x = 0

Solve : 6 – 10cosx = 3\(sin^2x\)

Find general solution of (2sinx – cosx)(1 + cosx) = \(sin^2x\)

Leave a Comment

Your email address will not be published.