Đáp án đúng: A
Giải chi tiết:\(\begin{array}{l}Pt \Leftrightarrow {\sin ^2}x - 1 - \sin x\cos x + \cos x + 2\sin 2x\left( {\sin x - 1} \right) = 0\\ \Leftrightarrow \left( {{{\sin }^2}x - 1} \right) - (\sin x\cos x - \cos x) + 2\sin 2x(\sin x - 1) = 0\\ \Leftrightarrow \left( {\sin x - 1} \right)\left( {\sin x + 1} \right) - \cos x\left( {\sin x - 1} \right) + 2\sin 2x(\sin x - 1) = 0\\ \Leftrightarrow \left( {\sin x - 1} \right)\left( {\sin x - \cos x + 2\sin 2x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\\\sin x - \cos x + 2\sin 2x + 1 = 0\,\,\,\left( 2 \right)\end{array} \right.\end{array}\)
Giải (1): \(x = \frac{\pi }{2} + k2\pi \left( {k \in \mathbb{Z}} \right).\)
Giải (2). Đặt \(\sin x - \cos x = \sqrt 2 \sin \left( {x - \frac{\pi }{4}} \right) = t\,\,\left( {t \in \left[ { - \sqrt 2 ;\sqrt 2 } \right]} \right) \Rightarrow \sin x\cos x = \frac{{1 - {t^2}}}{2}\)
\(\begin{array}{l} \Rightarrow \left( 2 \right) \Leftrightarrow \sin x - \cos x + 2\sin 2x + 1 = 0\\ \Leftrightarrow t + 4 \cdot \frac{{1 - {t^2}}}{2} + 1 = 0\\ \Leftrightarrow 2{t^2} - t - 3 = 0 \Leftrightarrow \left[ \begin{array}{l}t = - 1\;\;\;\left( {tm} \right)\\t = \frac{3}{2}\;\;\;\left( {ktm} \right)\end{array} \right.\end{array}\)
\(\begin{array}{l} \Rightarrow \sqrt 2 \sin \left( {x - \frac{\pi }{4}} \right) = - 1 \Leftrightarrow \sin \left( {x - \frac{\pi }{4}} \right) = \sin \left( { - \frac{\pi }{4}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x - \frac{\pi }{4} = - \frac{\pi }{4} + m2\pi \\x - \frac{\pi }{4} = \pi - \left( { - \frac{\pi }{4}} \right) + l2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = m2\pi \\x = \frac{{3\pi }}{2} + l2\pi \end{array} \right.\left( {m,\;l \in \mathbb{Z}} \right).\end{array}\)
Phương trình có nghiệm \(x \in \left( {0;\;6\pi } \right) \Leftrightarrow \left[ \begin{array}{l}0 < \frac{\pi }{2} + k2\pi < 6\pi \Leftrightarrow - \frac{1}{4} < k < \frac{{11}}{4} \Leftrightarrow k \in \left\{ {0;1;2} \right\}\\0 < \frac{{3\pi }}{2} + l2\pi < 6\pi \Leftrightarrow - \frac{3}{4} < l < \frac{9}{4} \Leftrightarrow l \in \left\{ {0;1;2} \right\}\\0 < m2\pi < 6\pi \Leftrightarrow 0 < m < 3 \Leftrightarrow m \in \left\{ {1;2} \right\}\end{array} \right.\)
Vậy phương trình có 8 nghiệm thõa mãn.
Chọn A.
