$3tan2x+3sin2x=cotx$ $(*)$
Điều kiện xác định $:$ $\left\{\begin{matrix} x \neq k\pi & \\ x \neq \frac{\pi}{4} +\frac{k\pi}{2} & \end{matrix}\right.$ $(k \in \mathbb{Z})$
$(*) \Leftrightarrow \frac{3sin2x}{cos2x}+3sin2x=\frac{cosx}{sinx} \Leftrightarrow 3sin2x(1+\frac{1}{cos2x})= \frac{cosx}{sinx} \Leftrightarrow \frac{3sin2x(cos2x+1)}{cos2x}= \frac{cosx}{sinx}$
$\Leftrightarrow 3sin2x.sinx(2cos^{2}x-1+1)=cos2x.cosx \Leftrightarrow 6sin2x.sinx.cos^{2}x=cosx(1-2sin^{2}x) \Leftrightarrow \left[\begin{matrix} cosx=0 & \\ 6sin2x.sinx.cosx=1-2sin^{2}x & (1) \end{matrix}\right.$
$(1) \Leftrightarrow 12sin^{2}x.cos^{2}x =1-2sin^{2}x \Leftrightarrow 12sin^{2}x.(1-sin^{2}x) =1-2sin^{2}x \Leftrightarrow 12sin^{2}x-12sin^{4}x=1-2sin^{2}x$
$\Leftrightarrow -12sin^{4}x+14sin^{2}x-1=0 \Leftrightarrow \cdots$