Câu hỏi 21

\(\begin{array}{l}\,\,\,\,\,\,\,{\rm{2sin}}x\left( {{\rm{1}} + {\rm{cos2}}x} \right) + \sin 2x = {\rm{1}} + {\rm{2}}\cos x\\ \Leftrightarrow 2\sin x.2{\cos ^2}x + \sin 2x - 1 - 2\cos x = 0\\ \Leftrightarrow 2\sin 2x.\cos x + \sin 2x - \left( {1 + 2\cos x} \right) = 0\\ \Leftrightarrow \sin 2x\left( {2\cos x + 1} \right) - \left( {1 + 2\cos x} \right) = 0\\ \Leftrightarrow \left( {\sin 2x - 1} \right)\left( {2\cos x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \frac{{ - 1}}{2}\\\sin 2x = 1\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x =  \pm \frac{{2\pi }}{3} + k\pi \\x = \frac{\pi }{4} + k\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

Vậy \(S = \left\{ { \pm \frac{{2\pi }}{3} + k2\pi ,\,\,\frac{\pi }{4} + k\pi ,\,\,k \in \mathbb{Z}} \right\}\).

Chọn B.