Giải phương trình: $\dfrac{1+sinx+cos2x)sin(x+\dfrac{π}{4})}{1+tanx}=\dfrac{1}{\sqrt{2}}cos x$

$\begin{array}{l} \dfrac{{\left( {1 + \sin x + \cos 2x} \right)\sin \left( {x + \dfrac{\pi }{4}} \right)}}{{1 + \tan x}} = \dfrac{1}{{\sqrt 2 }}\cos x\\ \text{ĐK}:\left\{ \begin{array}{l} \cos x \ne 0\\ 1 + \tan x \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ne \dfrac{\pi }{2} + k\pi \\ x \ne  - \dfrac{\pi }{4} + k\pi  \end{array} \right.\\  \Leftrightarrow \dfrac{{\left( {1 + \sin x + \cos 2x} \right)\dfrac{1}{{\sqrt 2 }}\left( {\sin x + \cos x} \right)}}{{1 + \tan x}} = \dfrac{1}{{\sqrt 2 }}\cos x\\  \Leftrightarrow \dfrac{{\left( {1 + \sin x + \cos 2x} \right)\left( {\sin x + \cos x} \right)}}{{1 + \tan x}} = \cos x\\  \Leftrightarrow \left( {1 + \sin x + \cos 2x} \right)\left( {\sin x + \cos x} \right) = \cos x\left( {1 + \tan x} \right)\\  \Leftrightarrow \left( {1 + \sin x + \cos 2x} \right)\left( {\sin x + \cos x} \right) = \cos x + \sin x\\  \Leftrightarrow \left( {\sin x + \cos x} \right)\left( {\sin x + \cos 2x + 1 - 1} \right) = 0\\  \Leftrightarrow \left[ \begin{array}{l} \sin x + \cos x = 0\\ \sin x + \cos 2x = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = 0\\  - 2{\sin ^2}x + \sin x + 1 = 0 \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} \sin \left( {x + \dfrac{\pi }{4}} \right) = 0\\ \sin x = 1\\ \sin x =  - \dfrac{1}{2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x =  - \dfrac{\pi }{4} + k\pi (L)\\ x = \dfrac{\pi }{2} + k2\pi (L)\\ x =  - \dfrac{\pi }{6} + k2\pi (tm)\\ x = \dfrac{{7\pi }}{6} + k2\pi (tm) \end{array} \right.\\  \Rightarrow S = \left\{ { - \dfrac{\pi }{6} + k2\pi ;\dfrac{{7\pi }}{6} + k2\pi ,k \in Z} \right\} \end{array}$