Đáp án đúng: D
Giải chi tiết:\(\begin{array}{l}\,\,\,\,\,\,2\sin 2x - \cos 2x = 7\sin x + 2\cos x - 4\\ \Leftrightarrow 4.\sin x.\cos x - 1 + 2{\sin ^2}x - 7\sin x - 2\cos x + 4 = 0\\ \Leftrightarrow \left( {4\sin x\cos x - 2\cos x} \right) + \left( {2{{\sin }^2}x - 7\sin x + 3} \right) = 0\\ \Leftrightarrow 2.\cos x.\left( {2\sin x - 1} \right) + \left( {2\sin x - 1} \right)\left( {\sin x - 3} \right) = 0\\ \Leftrightarrow \left( {2\sin x - 1} \right).\left( {2\cos x + \sin x - 3} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x = \frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\2\cos x + \sin x = 3\,\,\left( 2 \right)\end{array} \right.\end{array}\)
+) Xét \(\left( 1 \right) \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} + k2\pi \\x = \frac{{5\pi }}{6} + k2\pi \,\,\,\,\,\end{array} \right.\,\left( {k \in \mathbb{Z}} \right)\).
+) Xét \(\left( 2 \right):\,\,\sin x + 2\cos x = 3\).
Ta có: \(\sqrt {{A^2} + {B^2}} = \sqrt {{1^2} + {2^2}} = \sqrt 5 < \sqrt {{3^2}} = 3 \Rightarrow \) Phương trình (2) vô nghiệm.
Vậy \(S = \left\{ {\frac{\pi }{6} + k2\pi ;\frac{{5\pi }}{6} + k2\pi ,k \in \mathbb{Z}} \right\}\).
Chọn D.