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

Solution :

(2sinx – cosx)(1 + cosx) – (1 – \(cos^2x\)) = 0

\(\therefore\) (1 + cosx)(2sinx – cosx – 1 + cosx) = 0

\(\therefore\) (1 + cosx)(2sinx – 1) = 0

\(\implies\) cosx = -1 or sinx = \(1\over 2\)

\(\implies\) cosx = -1 = cos\(\pi\) \(\implies\) x = 2n\(\pi\) + \(\pi\) = (2n+1)\(\pi\), n \(\in\) I

or sinx = \(1\over 2\) = sin\(\pi\over 6\) \(\implies\) x = k\(\pi + (-1)^k{\pi\over 6}\), k \(\in\) I