Đáp án:

1) \(\left[ \begin{array}{l}x = \arcsin \left( {\dfrac{{1 - \sqrt 5 }}{2}} \right) + k2\pi \\x = \pi  - \arcsin \left( {\dfrac{{1 - \sqrt 5 }}{2}} \right) + k2\pi \end{array} \right.\,\,\left( {k \in Z} \right)\).

2) \(x = \dfrac{1}{2}\arctan \left( { - \dfrac{1}{{\sqrt 2 }}} \right) + \dfrac{{k\pi }}{2}\,\,\left( {k \in Z} \right)\).

Giải thích các bước giải:

1) \(\dfrac{{\left( {1 + \sin x + \cos 2x} \right)\sin \left( {x + \dfrac{\pi }{4}} \right)}}{{1 + \tan x}} = \dfrac{1}{{\sqrt 2 }}\cos x\)  (Đk: \(\tan x \ne  - 1 \Leftrightarrow x \ne  - \dfrac{\pi }{4} + k\pi \), \(\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi \))

\(\begin{array}{l} \Leftrightarrow \dfrac{{\left( {\sin x + 2{{\cos }^2}x} \right)\dfrac{1}{{\sqrt 2 }}\left( {\sin x + \cos x} \right)}}{{1 + \dfrac{{\sin x}}{{\cos x}}}} = \dfrac{1}{{\sqrt 2 }}\cos x\\ \Leftrightarrow \dfrac{{\left( {\sin x + 2{{\cos }^2}x} \right)\left( {\sin x + \cos x} \right)}}{{\dfrac{{\sin x + \cos x}}{{\cos x}}}} = \cos x\\ \Leftrightarrow \cos x\left( {\sin x + 2{{\cos }^2}x} \right) = \cos x\\ \Leftrightarrow \cos x\left( {\sin x + 1 - {{\sin }^2}x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = 0\,\,\left( {loai} \right)\\ - {\sin ^2}x + \sin x + 1 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\sin x = \dfrac{{1 + \sqrt 5 }}{2}\,\,\left( {loai} \right)\\\sin x = \dfrac{{1 - \sqrt 5 }}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \arcsin \left( {\dfrac{{1 - \sqrt 5 }}{2}} \right) + k2\pi \\x = \pi  - \arcsin \left( {\dfrac{{1 - \sqrt 5 }}{2}} \right) + k2\pi \end{array} \right.\,\,\left( {k \in Z} \right)\end{array}\)

Vậy \(\left[ \begin{array}{l}x = \arcsin \left( {\dfrac{{1 - \sqrt 5 }}{2}} \right) + k2\pi \\x = \pi  - \arcsin \left( {\dfrac{{1 - \sqrt 5 }}{2}} \right) + k2\pi \end{array} \right.\,\,\left( {k \in Z} \right)\).

2) \(\dfrac{{\cos x - \sqrt 2 \sin 2x}}{{2\cos x - 1}} = 1 + \cos x\)  (ĐK: \(\cos x \ne \dfrac{1}{2} \Leftrightarrow x \ne  \pm \dfrac{\pi }{3} + k2\pi \))

\(\begin{array}{l} \Leftrightarrow \cos x - \sqrt 2 \sin 2x = 2\cos x + 2{\cos ^2}x - 1 - \cos x\\ \Leftrightarrow  - \sqrt 2 \sin 2x = \cos 2x\\ \Leftrightarrow \tan 2x =  - \dfrac{1}{{\sqrt 2 }}\\ \Leftrightarrow 2x = \arctan \left( { - \dfrac{1}{{\sqrt 2 }}} \right) + k\pi \\ \Leftrightarrow x = \dfrac{1}{2}\arctan \left( { - \dfrac{1}{{\sqrt 2 }}} \right) + \dfrac{{k\pi }}{2}\,\,\left( {k \in Z} \right)\end{array}\)  

Vậy \(x = \dfrac{1}{2}\arctan \left( { - \dfrac{1}{{\sqrt 2 }}} \right) + \dfrac{{k\pi }}{2}\,\,\left( {k \in Z} \right)\).