ĐK: $\sin2x\ne 1\to x\ne \dfrac{\pi}{4}+k\pi$
$\dfrac{2\cos2x}{1-\sin2x}=0$
$\Rightarrow 2\cos2x=0$
$\Leftrightarrow \cos2x=0$
$\Leftrightarrow 2x=\dfrac{\pi}{2}+k\pi$
$\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}$
Đối chiếu ĐK, suy ra: $x=\dfrac{3\pi}{4}+k\pi$