Ta có:
$\cot^2x-\dfrac{\cos2x}{\sin^2x}=\dfrac{\cos^2x}{\sin^2x}-\dfrac{\cos^2x-\sin^2x}{\sin^2x}=1$
Đặt: $\cot^2x=u,1-\dfrac{\cos2x}{\sin^2x}=v$, ta được:
$\left\{\begin{array}{l}u+v=2\\4u^3+3v^4=7\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}u=2-v\\4(2-v)^3+3v^4=7\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}u=2-v\\(v-1)^2(3v^2+2v+25)=0\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}u=1\\v=1\end{array}\right.$
$\Leftrightarrow x=\pm\dfrac{\pi}{4}+k\pi,k\in\mathbb{Z}$