Phương trình 2sin x + cos x - sin 2x - 1 = 0 có nghiệm là:
A. $\left[ \begin{array}{l}x = \frac{\pi }{6} + k\pi \\x = \frac{{5\pi }}{6} + k\pi \\x = k\pi \end{array} \right.$, $k \in \mathbb{Z}$.
B. $\left[ \begin{array}{l}x = \frac{\pi }{6} + k2\pi \\x = \frac{{5\pi }}{6} + k2\pi \\x = k2\pi \end{array} \right.$, $k \in \mathbb{Z}$.
C. $\left[ \begin{array}{l}x = \frac{\pi }{6} + k2\pi \\x = - \frac{\pi }{6} + k2\pi \\x = k2\pi \end{array} \right.$, $k \in \mathbb{Z}$.
D. $\left[ \begin{array}{l}x = \frac{\pi }{6} + k2\pi \\x = - \frac{\pi }{6} + k2\pi \\x = k\pi \end{array} \right.$, $k \in \mathbb{Z}$.
A. $\left[ \begin{array}{l}x = \frac{\pi }{6} + k\pi \\x = \frac{{5\pi }}{6} + k\pi \\x = k\pi \end{array} \right.$, $k \in \mathbb{Z}$.
B. $\left[ \begin{array}{l}x = \frac{\pi }{6} + k2\pi \\x = \frac{{5\pi }}{6} + k2\pi \\x = k2\pi \end{array} \right.$, $k \in \mathbb{Z}$.
C. $\left[ \begin{array}{l}x = \frac{\pi }{6} + k2\pi \\x = - \frac{\pi }{6} + k2\pi \\x = k2\pi \end{array} \right.$, $k \in \mathbb{Z}$.
D. $\left[ \begin{array}{l}x = \frac{\pi }{6} + k2\pi \\x = - \frac{\pi }{6} + k2\pi \\x = k\pi \end{array} \right.$, $k \in \mathbb{Z}$.
Chọn B.
$2\sin x + \cos x - \sin 2x - 1 = 0 \Leftrightarrow 2\sin x + \cos x - 2\sin x\cos x - 1 = 0$
$ \Leftrightarrow \left( {\cos x - 1} \right)\left( {1 - 2\sin x} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}\cos x = 1\\\sin x = \frac{1}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} + k2\pi \\x = \frac{{5\pi }}{6} + k2\pi \\x = k2\pi \end{array} \right.$
$2\sin x + \cos x - \sin 2x - 1 = 0 \Leftrightarrow 2\sin x + \cos x - 2\sin x\cos x - 1 = 0$
$ \Leftrightarrow \left( {\cos x - 1} \right)\left( {1 - 2\sin x} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}\cos x = 1\\\sin x = \frac{1}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} + k2\pi \\x = \frac{{5\pi }}{6} + k2\pi \\x = k2\pi \end{array} \right.$