Solve the equation: $\;$ $\tan 2x \sin x + \sqrt{3} \left(\sin x - \sqrt{3} \tan 2x\right) = 3 \sqrt{3}$
Given equation: $\;\;\;$ $\tan 2x \sin x + \sqrt{3} \left(\sin x - \sqrt{3} \tan 2x\right) = 3 \sqrt{3}$
i.e. $\;$ $\tan 2x \sin x + \sqrt{3} \sin x - 3 \tan 2x - 3 \sqrt{3} = 0$
i.e. $\;$ $\tan 2x \left(\sin x - 3\right) + \sqrt{3} \left(\sin x - 3\right) = 0$
i.e. $\;$ $\left(\sin x - 3\right) \left(\tan 2x + \sqrt{3}\right) = 0$
$\implies$ $\sin x - 3 = 0$ $\;$ or $\tan 2x + \sqrt{3} = 0$
Case 1:
$\sin x - 3 = 0$
i.e. $\;$ $\sin x = 3$
But sine of any angle cannot be greater than $1$.
$\therefore \;$ $\sin x = 3$ does not give a valid solution.
Case 2:
$\tan 2x + \sqrt{3} = 0$
i.e. $\;$ $\tan 2x = - \sqrt{3}$
i.e. $\;$ $2x = n \pi + \tan^{-1} \left(- \sqrt{3}\right)$
i.e. $\;$ $2x = n \pi - \tan^{-1} \left(\sqrt{3}\right)$
i.e. $\;$ $2x = n \pi - \dfrac{\pi}{3}$
$\therefore \;$ $x = \dfrac{n \pi}{2} - \dfrac{\pi}{6}, \;\;\; n \in Z$