Điều kiện xác định $\cos x\ne 0\Rightarrow x\ne \dfrac{\pi} 2+k\pi $
$\begin{array}{l} \dfrac{{1 + \sin 2x - \cos 2x}}{{1 + {{\tan }^2}x}} = \cos x\left( {\sin 2x + 2{{\cos }^2}x} \right)\\ \Leftrightarrow \dfrac{{1 + \sin 2x - \left( {1 - 2{{\sin }^2}x} \right)}}{{\dfrac{1}{{{{\cos }^2}x}}}} = \cos x\left( {2\sin x\cos x + 2{{\cos }^2}x} \right)\\ \Leftrightarrow \dfrac{{2{{\sin }^2}x + 2\sin x\cos x}}{{\dfrac{1}{{{{\cos }^2}x}}}} = 2{\cos ^2}x\left( {\sin x + \cos x} \right)\\ \Leftrightarrow 2\sin x\left( {\sin x + \cos x} \right).{\cos ^2}x = 2{\cos ^2}x\left( {\sin x + \cos x} \right)\\ \Leftrightarrow \sin x\left( {\sin x + \cos x} \right) = \left( {\sin x + \cos x} \right)\\ \Leftrightarrow \left( {\sin x + \cos x} \right)\left( {\sin x - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin x + \cos x = 0\\ \sin x = 1 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = 0\\ x = \dfrac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x + \dfrac{\pi }{4} = k\pi \\ x = \dfrac{\pi }{2} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - \dfrac{\pi }{4} + k\pi \\ x = \dfrac{\pi }{2} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right) \end{array}$
