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