Solve the equation: $\;$ $\left(2 \sin x - \cos x\right) \left(1 + \cos x\right) = \sin^2 x$
Given equation: $\;\;\;$ $\left(2 \sin x - \cos x\right) \left(1 + \cos x\right) = \sin^2 x$
i.e. $\;$ $2 \sin x + 2 \sin x \cos x - \cos x - \cos^2 x = \sin^2 x$
i.e. $\;$ $2 \sin x + 2 \sin x \cos x - \cos x = \sin^2 x + \cos^2 x$
i.e. $\;$ $2 \sin x + 2 \sin x \cos x - \cos x = 1$
i.e. $\;$ $2 \sin x \left(1 + \cos x\right) - \left(1 + \cos x\right) = 0$
i.e. $\;$ $\left(2 \sin x - 1\right) \left(1 + \cos x\right) = 0$
i.e. $\;$ $2 \sin x - 1 = 0$ $\;\;\;$ or $\;\;\;$ $1 + \cos x = 0$
When $\;\;$ $2 \sin x - 1 = 0$
i.e. $\;$ $\sin x = \dfrac{1}{2}$
$\implies$ $x = \left(-1\right)^n \sin^{-1} \left(\dfrac{1}{2}\right) + n \pi, \;\;\; n \in Z$
i.e. $\;$ $x = \left(-1\right)^n \times \dfrac{\pi}{6} + n \pi, \;\;\; n \in Z$
Or when $\;\;$ $1 + \cos x = 0$
i.e. $\;$ $\cos x = -1$
$\implies$ $x = \pm \cos^{-1} \left(-1\right) + 2 m \pi, \;\;\; m \in Z$
i.e. $\;$ $x = 2 m \pi \pm \pi, \;\;\; m \in Z$