ĐK: $\cos2x\ne 0\to x\ne \dfrac{\pi}{4}+\dfrac{k\pi}{2}$
$\sin x\tan2x-\tan2x=\sin x-1$
$\to \tan2x(\sin x-1)=\sin x-1$
$\to (\tan2x-1)(\sin x-1)=0$
$\to \left[ \begin{array}{l}\tan2x=1 \\ \sin x=1\end{array} \right.$
$\to \left[ \begin{array}{l}x=\dfrac{\pi}{8}+\dfrac{k\pi}{2} \quad(TM) \\x=\dfrac{\pi}{2}+k2\pi \quad(TM)\end{array} \right.$
