Phương trình $\left( {2\sin x + 1} \right)\left( {3\cos 4x + 2\sin x - 4} \right) + 4{\cos ^2}x = 3$ có nghiệm là:
A. $\left[ \begin{array}{l}x = - \frac{\pi }{6} + k2\pi \\x = \frac{{7\pi }}{6} + k2\pi \\x = k\frac{\pi }{2}\end{array} \right.$.
B. $\left[ \begin{array}{l}x = \frac{\pi }{6} + k2\pi \\x = \frac{{5\pi }}{6} + k2\pi \\x = k\pi \end{array} \right.$.
C. $\left[ \begin{array}{l}x = - \frac{\pi }{3} + k2\pi \\x = \frac{{4\pi }}{3} + k2\pi \\x = k2\pi \end{array} \right.$.
D. $\left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x = \frac{{2\pi }}{3} + k2\pi \\x = k\frac{{2\pi }}{3}\end{array} \right.$.
A. $\left[ \begin{array}{l}x = - \frac{\pi }{6} + k2\pi \\x = \frac{{7\pi }}{6} + k2\pi \\x = k\frac{\pi }{2}\end{array} \right.$.
B. $\left[ \begin{array}{l}x = \frac{\pi }{6} + k2\pi \\x = \frac{{5\pi }}{6} + k2\pi \\x = k\pi \end{array} \right.$.
C. $\left[ \begin{array}{l}x = - \frac{\pi }{3} + k2\pi \\x = \frac{{4\pi }}{3} + k2\pi \\x = k2\pi \end{array} \right.$.
D. $\left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x = \frac{{2\pi }}{3} + k2\pi \\x = k\frac{{2\pi }}{3}\end{array} \right.$.
Chọn A
$\left( {2\sin x + 1} \right)\left( {3\cos 4x + 2\sin x - 4} \right) + 4{\cos ^2}x = 3$
$ \Leftrightarrow \left( {2\sin x + 1} \right)\left( {3\cos 4x + 2\sin x - 4} \right) + 4\left( {1 - {{\sin }^2}x} \right) - 3 = 0$
$ \Leftrightarrow \left( {2\sin x + 1} \right)\left( {3\cos 4x + 2\sin x - 4} \right) + \left( {1 - 4{{\sin }^2}x} \right) = 0$
$ \Leftrightarrow \left( {2\sin x + 1} \right)\left( {3\cos 4x + 2\sin x - 4 + 1 - 2\sin x} \right) = 0$
$ \Leftrightarrow \left( {2\sin x + 1} \right)\left( {3\cos 4x - 3} \right) = 0$$ \Leftrightarrow \left[ \begin{array}{l}\sin x = - \frac{1}{2}\\\cos 4x = 1\end{array} \right.$$ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{6} + k2\pi \\x = \frac{{7\pi }}{6} + k2\pi \\x = k\frac{\pi }{2}\end{array} \right.,\,\,\left( {k \in \mathbb{Z}} \right)$
$\left( {2\sin x + 1} \right)\left( {3\cos 4x + 2\sin x - 4} \right) + 4{\cos ^2}x = 3$
$ \Leftrightarrow \left( {2\sin x + 1} \right)\left( {3\cos 4x + 2\sin x - 4} \right) + 4\left( {1 - {{\sin }^2}x} \right) - 3 = 0$
$ \Leftrightarrow \left( {2\sin x + 1} \right)\left( {3\cos 4x + 2\sin x - 4} \right) + \left( {1 - 4{{\sin }^2}x} \right) = 0$
$ \Leftrightarrow \left( {2\sin x + 1} \right)\left( {3\cos 4x + 2\sin x - 4 + 1 - 2\sin x} \right) = 0$
$ \Leftrightarrow \left( {2\sin x + 1} \right)\left( {3\cos 4x - 3} \right) = 0$$ \Leftrightarrow \left[ \begin{array}{l}\sin x = - \frac{1}{2}\\\cos 4x = 1\end{array} \right.$$ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{6} + k2\pi \\x = \frac{{7\pi }}{6} + k2\pi \\x = k\frac{\pi }{2}\end{array} \right.,\,\,\left( {k \in \mathbb{Z}} \right)$