Đáp án đúng: B
Giải chi tiết:ĐK : \(\sin 2x \ne \frac{1}{2} \Leftrightarrow \left\{ \begin{array}{l}2x \ne \frac{{5\pi }}{6} + k2\pi \\2x \ne \frac{\pi }{6} + m2\pi \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne \frac{{5\pi }}{{12}} + k2\pi \\x \ne \frac{\pi }{{12}} + m2\pi \end{array} \right.\left( {m,\;k \in \mathbb{Z}} \right)\)
\(\begin{array}{l}\sin 3x - \cos 3x = 3\sin x - 4{\sin ^3}x - 4{\cos ^3}x + 3\cos x\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\;\;\;\;\; = 3\left( {\sin x + \cos x} \right) - 4\left( {{{\sin }^3}x + {{\cos }^3}x} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\; = 3(\sin x + \cos x) - 4(\sin x + \cos x)(1 - \sin x\cos x)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\,\,\;\; = (\sin x + \cos x)(4\sin x\cos x - 1) = (\sin x + \cos x)(2\sin 2x - 1)\\ \Rightarrow \frac{{\sin 3x - \cos 3x}}{{2\sin 2x - 1}} = \sin x + \cos x.\\Pt \Leftrightarrow 7(\sin x + \cos x - \cos x) = 4 - \cos 2x \Leftrightarrow 7\sin x = 4 - (1 - 2{\sin ^2}x)\\ \Leftrightarrow 2{\sin ^2}x - 7\sin x + 3 = 0 \Leftrightarrow \left[ \begin{array}{l}\sin x = \frac{1}{2}\;\;\left( {tm} \right)\\\sin \;x = 3\;\;\;\left( {ktm} \right)\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} + k2\pi \\x = \frac{{5\pi }}{6} + l2\pi \end{array} \right.\;\;\left( {k,\;l \in Z} \right)\end{array}\)
Chọn nghiệm trên khoảng \(\left( {0;\pi } \right)\) ta được hai nghiệm của phương trình là:m \(x = \frac{\pi }{6};\;\;x = \frac{{5\pi }}{6}.\)
Chọn B.
