ĐKXĐ: \(sin2x\ne1\)
\(\dfrac{2cos2x}{1-sin2x}=0\)
\(\Rightarrow cos2x=0\)
\(\Leftrightarrow cos^22x=0\)
\(\Leftrightarrow1-sin^22x=0\)
\(\Rightarrow\left[{}\begin{matrix}sin2x=1\left(loại\right)\\sin2x=-1\end{matrix}\right.\)
\(\Rightarrow x=-\dfrac{\pi}{4}+k\pi\)