Solve the equation: $\;$ $6 \cos^2 x + 5 \sin x - 7 = 0$
Given equation: $\;\;\;$ $6 \cos^2 x + 5 \sin x - 7 = 0$
i.e. $\;$ $6 \left(1 - \sin^2 x\right) + 5 \sin x - 7 = 0$
i.e. $\;$ $6 - 6 \sin^2 x + 5 \sin x - 7 = 0$
i.e. $\;$ $6 \sin^2 x - 5 \sin x + 1 = 0$
i.e. $\;$ $6 \sin^2 x - 3 \sin x - 2 \sin x + 1 = 0$
i.e. $\;$ $3 \sin x \left(2 \sin x - 1\right) - \left(2 \sin x - 1\right) = 0$
i.e. $\;$ $\left(3 \sin x - 1\right) \left(2 \sin x - 1\right) = 0$
i.e. $\;$ $3 \sin x - 1 = 0$ $\;$ or $\;$ $2 \sin x - 1 = 0$
Case 1: $\;$ $3 \sin x - 1 = 0$
i.e. $\;$ $\sin x = \dfrac{1}{3}$
i.e. $\;$ $x = n \pi + \left(-1\right)^n \sin^{-1} \left(\dfrac{1}{3}\right), \;\;\; n \in Z$
Case 2: $\;$ $2 \sin x - 1 = 0$
i.e. $\;$ $\sin x = \dfrac{1}{2} = \sin \left(\dfrac{\pi}{6}\right)$
i.e. $\;$ $x = m \pi + \left(-1\right)^m \times \dfrac{\pi}{6}, \;\;\; m \in Z$