Giải
ĐK: $x \neq \dfrac{\pi}{2} + k\pi \, (k \in Z)$
Phương trình ban đầu tương đương:
$(3 + \cos{2x})\tan{x} = \sqrt{3}\cos{2x}\tan^2{x} + \sin{2x} + \sqrt{3}\cos{2x}$
$\Leftrightarrow \left (3 + \dfrac{1 - \tan^2{x}}{1 + \tan^2{x}} \right)\tan{x} = \sqrt{3}\cos{2x}(1 + \tan^2{x}) + \dfrac{2\tan{x}}{1 + \tan^2{x}}$
$\Leftrightarrow \dfrac{4 + 2\tan^2{x}}{1 + \tan^2{x}}\tan{x} - \dfrac{2\tan{x}}{1 + \tan^2{x}} = \sqrt{3}(1 - \tan^2{x})$
$\Leftrightarrow 2\tan{x} = \sqrt{3}(1 - \tan^2{x}) \Leftrightarrow \left[\begin{matrix}\tan{x} = - \sqrt{3}\\\tan{x} = \dfrac{1}{\sqrt{3}}\end{matrix}\right. \Leftrightarrow \left[\begin{matrix}x = \dfrac{-\pi}{3} + k\pi\\x = \dfrac{\pi}{6} + k\pi\end{matrix}\right. \, (k \in Z)$