Phương trình $2\sin \left( {3x + \frac{\pi }{4}} \right) = \sqrt {1 + 8\sin 2x.{{\cos }^2}2x} $ có nghiệm là:
A. $\left[ \begin{array}{l}x = \frac{\pi }{6} + k\pi \\x = \frac{{5\pi }}{6} + k\pi \end{array} \right.$.
B. $\left[ \begin{array}{l}x = \frac{\pi }{{12}} + k\pi \\x = \frac{{5\pi }}{{12}} + k\pi \end{array} \right.$.
C. $\left[ \begin{array}{l}x = \frac{\pi }{{18}} + k\pi \\x = \frac{{5\pi }}{{18}} + k\pi \end{array} \right.$.
D. $\left[ \begin{array}{l}x = \frac{\pi }{{24}} + k\pi \\x = \frac{{5\pi }}{{24}} + k\pi \end{array} \right.$.
A. $\left[ \begin{array}{l}x = \frac{\pi }{6} + k\pi \\x = \frac{{5\pi }}{6} + k\pi \end{array} \right.$.
B. $\left[ \begin{array}{l}x = \frac{\pi }{{12}} + k\pi \\x = \frac{{5\pi }}{{12}} + k\pi \end{array} \right.$.
C. $\left[ \begin{array}{l}x = \frac{\pi }{{18}} + k\pi \\x = \frac{{5\pi }}{{18}} + k\pi \end{array} \right.$.
D. $\left[ \begin{array}{l}x = \frac{\pi }{{24}} + k\pi \\x = \frac{{5\pi }}{{24}} + k\pi \end{array} \right.$.
Chọn B
Điều kiện $1 + 8\sin 2x.{\cos ^2}2x \ge 0$
$2\sin \left( {3x + \frac{\pi }{4}} \right) = \sqrt {1 + 8\sin 2x.{{\cos }^2}2x} $$ \Leftrightarrow 4{\sin ^2}\left( {3x + \frac{\pi }{4}} \right) = 1 + 8\sin 2x.{\cos ^2}2x$.
$ \Leftrightarrow 2\left[ {1 - \cos \left( {6x + \frac{\pi }{2}} \right)} \right] = 1 + 8\sin 2x.{\cos ^2}2x \Leftrightarrow 8\sin 2x.{\cos ^2}2x - 2\sin 6x - 1 = 0$.
$ \Leftrightarrow 8\sin 2x\left( {1 - {{\sin }^2}2x} \right) - 2\left( {3\sin 2x - 4{{\sin }^3}2x} \right) - 1 = 0 \Leftrightarrow 2\sin 2x - 1 = 0$
$ \Leftrightarrow \sin 2x = \frac{1}{2} \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{12}} + k\pi \\x = \frac{{5\pi }}{{12}} + k\pi \end{array} \right.$, $k \in \mathbb{Z}$.
Thử lại điều kiện, $\left[ \begin{array}{l}x = \frac{\pi }{{12}} + k\pi \\x = \frac{{5\pi }}{{12}} + k\pi \end{array} \right.$, $k \in \mathbb{Z}$ đều thoả.
Điều kiện $1 + 8\sin 2x.{\cos ^2}2x \ge 0$
$2\sin \left( {3x + \frac{\pi }{4}} \right) = \sqrt {1 + 8\sin 2x.{{\cos }^2}2x} $$ \Leftrightarrow 4{\sin ^2}\left( {3x + \frac{\pi }{4}} \right) = 1 + 8\sin 2x.{\cos ^2}2x$.
$ \Leftrightarrow 2\left[ {1 - \cos \left( {6x + \frac{\pi }{2}} \right)} \right] = 1 + 8\sin 2x.{\cos ^2}2x \Leftrightarrow 8\sin 2x.{\cos ^2}2x - 2\sin 6x - 1 = 0$.
$ \Leftrightarrow 8\sin 2x\left( {1 - {{\sin }^2}2x} \right) - 2\left( {3\sin 2x - 4{{\sin }^3}2x} \right) - 1 = 0 \Leftrightarrow 2\sin 2x - 1 = 0$
$ \Leftrightarrow \sin 2x = \frac{1}{2} \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{12}} + k\pi \\x = \frac{{5\pi }}{{12}} + k\pi \end{array} \right.$, $k \in \mathbb{Z}$.
Thử lại điều kiện, $\left[ \begin{array}{l}x = \frac{\pi }{{12}} + k\pi \\x = \frac{{5\pi }}{{12}} + k\pi \end{array} \right.$, $k \in \mathbb{Z}$ đều thoả.