\[\begin{array}{l}
\frac{{2\cos 2x}}{{1 - \sin 2x}} = 0\\
DK:1 - \sin 2x \ne 0 \Leftrightarrow \sin 2x \ne 1\\
PT \Leftrightarrow \cos 2x = 0 \Leftrightarrow {\sin ^2}2x = 1 \Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 1\left( {Loai\,do\,DK} \right)\\
\sin 2x = - 1
\end{array} \right.\\
\Leftrightarrow \sin 2x = - 1 \Leftrightarrow 2x = - \frac{\pi }{2} + k2\pi \Leftrightarrow x = - \frac{\pi }{4} + k\pi
\end{array}\]
