Phương trình: $\sin x.\cos 4x - {\sin ^2}2x = 4{\sin ^2}\left( {\frac{\pi }{4} - \frac{x}{2}} \right) - \frac{7}{2}$có nghiệm là
A. $\left[ \begin{array}{l}x = - \frac{\pi }{6} + k\pi \\x = \frac{{7\pi }}{6} + k\pi \end{array} \right.$, $k \in \mathbb{Z}$.
B. $\left[ \begin{array}{l}x = - \frac{\pi }{6} + k2\pi \\x = \frac{{7\pi }}{6} + k2\pi \end{array} \right.$, $k \in \mathbb{Z}$.
C. $\left[ \begin{array}{l}x = - \frac{\pi }{6} + k2\pi \\x = \frac{\pi }{6} + k2\pi \end{array} \right.$, $k \in \mathbb{Z}$.
D. $\left[ \begin{array}{l}x = - \frac{\pi }{6} + k\pi \\x = \frac{\pi }{6} + k\pi \end{array} \right.$, $k \in \mathbb{Z}$.
A. $\left[ \begin{array}{l}x = - \frac{\pi }{6} + k\pi \\x = \frac{{7\pi }}{6} + k\pi \end{array} \right.$, $k \in \mathbb{Z}$.
B. $\left[ \begin{array}{l}x = - \frac{\pi }{6} + k2\pi \\x = \frac{{7\pi }}{6} + k2\pi \end{array} \right.$, $k \in \mathbb{Z}$.
C. $\left[ \begin{array}{l}x = - \frac{\pi }{6} + k2\pi \\x = \frac{\pi }{6} + k2\pi \end{array} \right.$, $k \in \mathbb{Z}$.
D. $\left[ \begin{array}{l}x = - \frac{\pi }{6} + k\pi \\x = \frac{\pi }{6} + k\pi \end{array} \right.$, $k \in \mathbb{Z}$.
Chọn B
$\sin x.\cos 4x - \frac{{1 - \cos 4x}}{2} = 2\left( {1 - \sin x} \right) - \frac{7}{2}$$ \Leftrightarrow \cos 4x\left( {\sin x + \frac{1}{2}} \right) = - 2\left( {\sin x + \frac{1}{2}} \right)$
$ \Leftrightarrow \left( {\sin x + \frac{1}{2}} \right)\left( {\cos 4x + 2} \right) = 0$$ \Leftrightarrow \sin x = - \frac{1}{2} \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{6} + k2\pi \\x = \frac{{7\pi }}{6} + k2\pi \end{array} \right.$, $k \in \mathbb{Z}$
$\sin x.\cos 4x - \frac{{1 - \cos 4x}}{2} = 2\left( {1 - \sin x} \right) - \frac{7}{2}$$ \Leftrightarrow \cos 4x\left( {\sin x + \frac{1}{2}} \right) = - 2\left( {\sin x + \frac{1}{2}} \right)$
$ \Leftrightarrow \left( {\sin x + \frac{1}{2}} \right)\left( {\cos 4x + 2} \right) = 0$$ \Leftrightarrow \sin x = - \frac{1}{2} \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{6} + k2\pi \\x = \frac{{7\pi }}{6} + k2\pi \end{array} \right.$, $k \in \mathbb{Z}$