Nghiệm phương trình \(\left( {1 - {\rm{tan}}x} \right)\left( {1 + {\rm{sin}}2x} \right) = 1 + {\rm{tan}}x\) là    

Điều kiện: \({\rm{cos}}x \ne 0 \Leftrightarrow x \ne \frac{\pi }{2} + k\pi \)

\(PT \Leftrightarrow \frac{{\cos x - \sin x}}{{\cos x}}{\left( {\sin x + \cos x} \right)^2} = \frac{{\cos x + \sin x}}{{\cos x}}\)

\[ \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{\sin x + \cos x = 0}\\{\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right) = 1}\end{array}} \right.\]

\( \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{\sin x + \cos x = 0}\\{{{\cos }^2}x - {{\sin }^2}x = {{\sin }^2}x + {{\cos }^2}x}\end{array}} \right.\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{ sin x +  cos x = 0}\\{ sin x = 0}\end{array} \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{\sqrt 2  sin \left( {x + \frac{\pi }{4}} \right) = 0}\\{ sin x = 0}\end{array} \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{x =  - \frac{\pi }{4} + k\pi }\\{x = k\pi }\end{array}} \right.} \right.} \right.,k \in \mathbb{Z}\).  Chọn D.