Solve the equation: $\;$ $2 \sin x + \tan x = 0$
Given equation: $\;\;\;$ $2 \sin x + \tan x = 0$
i.e. $\;$ $2 \sin x + \dfrac{\sin x}{\cos x} = 0$
i.e. $\;$ $2 \sin x \cos x + \sin x = 0$
i.e. $\;$ $\sin x \left(2 \cos x + 1\right) = 0$
$\implies$ $\sin x = 0$ $\;\;\;$ or $\;\;\;$ $2 \cos x + 1 = 0$
When $\;\;$ $\sin x = 0$
$\implies$ $x = n \pi, \;\;\; n \in I$
When $\;\;$ $2 \cos x + 1 = 0$
$\implies$ $\cos x = \dfrac{-1}{2} = \cos \left(\pi - \dfrac{\pi}{3}\right) = \cos \left(\dfrac{2 \pi}{3}\right)$
i.e. $\;$ $x = 2 m \pi \pm \dfrac{2 \pi}{3}, \;\;\; m \in I$