Solve the equation: $\;$ $\cos 2x + 3 \sin x = 2$
Given equation: $\;\;\;$ $\cos 2x + 3 \sin x = 2$
i.e. $\;$ $1 - 2 \sin^2 x + 3 \sin x = 2$
i.e. $\;$ $2 \sin^2 x - 3 \sin x + 1 = 0$
i.e. $\;$ $2 \sin^2 x - 2 \sin x - \sin x + 1 = 0$
i.e. $\;$ $\left(2 \sin x - 1\right) \left(\sin x - 1\right) = 0$
i.e. $\;$ $2 \sin x - 1 = 0$ $\;$ or $\;$ $\sin x - 1 = 0$
Case 1: $\;$ $2 \sin x - 1 = 0$
i.e. $\;$ $\sin x = \dfrac{1}{2} = \sin \left(\dfrac{\pi}{6}\right)$
i.e. $\;$ $x = n \pi + \left(-1\right)^n \times \dfrac{\pi}{6}, \;\;\; n \in Z$
Case 2: $\;$ $\sin x - 1 = 0$
i.e. $\;$ $\sin x = 1 = \sin \left(\dfrac{\pi}{2}\right)$
i.e. $\;$ $x = 2 m \pi + \dfrac{\pi}{2}, \;\;\; m \in Z$