Câu hỏi 20

\(\begin{array}{l}\,\,\,\,\,\,\,\sqrt 2 \left( {\sin x - 2\cos x} \right) = 2 - \sin 2x\\ \Leftrightarrow \sqrt 2 \left( {\sin x - 2\cos x} \right) = 2 - 2\sin x\cos x\\ \Leftrightarrow \sqrt 2 \sin x + 2\sin x.\cos x - 2\sqrt 2 \cos x - 2 = 0\\ \Leftrightarrow \sqrt 2 \sin x\left( {1 + \sqrt 2 \cos x} \right) - 2\left( {\sqrt 2 \cos x + 1} \right) = 0\\ \Leftrightarrow \left( {\sqrt 2 \sin x - 2} \right)\left( {1 + \sqrt 2 \cos x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sqrt 2 \sin x - 2 = 0\\1 + \sqrt 2 \cos x = 0\end{array} \right. \Rightarrow \left[ \begin{array}{l}\sin x = \sqrt 2 \,\,\,\left( {loai} \right)\\\cos x = \frac{{ - 1}}{{\sqrt 2 }} = \cos \frac{{3\pi }}{4}\end{array} \right. \Leftrightarrow x =  \pm \frac{{3\pi }}{4} + 2k\pi \,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

Vậy \(S = \left\{ { \pm \frac{{3\pi }}{4} + k2\pi ,\,\,k \in \mathbb{Z}} \right\}\).

Chọn B.