Em tách ra hỏi từng ý một thôi nhé
\(\begin{array}{l}
1)\,DK:\,\,\cos x \ne 0\\
\left( {1 - \tan x} \right)\left( {1 + \sin 2x} \right) = 1 + \tan x\\
\Leftrightarrow \left( {1 - \dfrac{{\sin x}}{{\cos x}}} \right)\left( {{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x} \right) = 1 + \dfrac{{\sin x}}{{\cos x}}\\
\Leftrightarrow \dfrac{{\cos x - \sin x}}{{\cos x}}{\left( {\cos x + \sin x} \right)^2} = \dfrac{{\cos x + \sin x}}{{\cos x}}\\
\Leftrightarrow \left( {\cos x + \sin x} \right)\left( {{{\cos }^2}x - {{\sin }^2}x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x + \sin x = 0\\
\cos 2x = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos \left( {x - \dfrac{\pi }{4}} \right) = 0\\
\cos 2x = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{3\pi }}{4} + k\pi \\
x = k\pi
\end{array} \right.
\end{array}\)
