Đặt $\tan x=t$, phương trình trở thành:
$6t+\frac{5(1-3t^2)}{3t-t^3}=\frac{2t}{1-t^2}$
$\Leftrightarrow 6t^6-7t^4-8t^2+5=0$
$\Leftrightarrow (t^2+1)(3t^2-5)(2t^2-1)=0$
$\Leftrightarrow \left[\begin{array}{l}t=\pm\sqrt{\frac{5}{3}}\\t=\pm\sqrt{\frac{1}{2}}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x=\pm\arctan\sqrt{\frac{5}{3}}+k\pi\\x=\pm\arctan\sqrt{\frac{1}{2}}+k\pi\end{array}\right.,k\in\mathbb{Z}$