Giải phương trình \(\sin 3x + \cos 2x - \sin x = 0...

B

- Hướng dẫn giải

Lời giải chi tiết:

\(\begin{array}{l}\,\,\,\,\,\,\,\sin 3x + \cos 2x - \sin x = 0\\ \Leftrightarrow 3\sin x - 4{\sin ^3}x + 1 - 2{\sin ^2}x - \sin x = 0\\ \Leftrightarrow  - 4{\sin ^3}x - 2{\sin ^2}x + 2\sin x + 1 = 0\\ \Leftrightarrow  - 2{\sin ^2}x\left( {2\sin x + 1} \right) + \left( {2\sin x + 1} \right) = 0\\ \Leftrightarrow \left( {2\sin x + 1} \right)\left( {1 - 2{{\sin }^2}x} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}2\sin x + 1 = 0\\1 - 2{\sin ^2}x = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\sin x = \frac{{ - 1}}{2}\\\cos 2x = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{ - \pi }}{6} + k2\pi \\x = \frac{{7\pi }}{6} + k2\pi \\x = \frac{\pi }{4} + \frac{{k\pi }}{2}\end{array} \right.\,\,\,\,\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

Vậy \(S = \left\{ {\frac{{ - \pi }}{6} + k2\pi ;\frac{{7\pi }}{6} + k2\pi ,\frac{\pi }{4} + \frac{{k\pi }}{2},k \in \mathbb{Z}} \right\}\).

Chọn B.