Phương trình: $\left( {\sin x - \sin 2x} \right)\left( {\sin x + \sin 2x} \right) = {\sin ^2}3x$ có các nghiệm là:
A. $\left[ \begin{array}{l}x = k\frac{\pi }{3}\\x = k\frac{\pi }{2}\end{array} \right.$.
B. $\left[ \begin{array}{l}x = k\frac{\pi }{6}\\x = k\frac{\pi }{4}\end{array} \right.$.
C. $\left[ \begin{array}{l}x = k\frac{{2\pi }}{3}\\x = k\pi \end{array} \right.$.
D. $\left[ \begin{array}{l}x = k3\pi \\x = k2\pi \end{array} \right.$.
A. $\left[ \begin{array}{l}x = k\frac{\pi }{3}\\x = k\frac{\pi }{2}\end{array} \right.$.
B. $\left[ \begin{array}{l}x = k\frac{\pi }{6}\\x = k\frac{\pi }{4}\end{array} \right.$.
C. $\left[ \begin{array}{l}x = k\frac{{2\pi }}{3}\\x = k\pi \end{array} \right.$.
D. $\left[ \begin{array}{l}x = k3\pi \\x = k2\pi \end{array} \right.$.
Chọn A
$\left( {\sin x - \sin 2x} \right)\left( {\sin x + \sin 2x} \right) = {\sin ^2}3x$ $ \Leftrightarrow {\sin ^2}x - {\sin ^2}2x = {\sin ^2}3x$.
$ \Leftrightarrow \frac{{1 - \cos 2x}}{2} - {\sin ^2}2x = \frac{{1 - \cos 6x}}{2}$ $ \Leftrightarrow \cos 6x - \cos 2x - 2{\sin ^2}2x = 0$
$ \Leftrightarrow - 2\cos 4x.\sin 2x - 2{\sin ^2}2x = 0$ $ \Leftrightarrow 2{\sin ^2}2x.\cos 2x + {\sin ^2}2x = 0$.
$ \Leftrightarrow {\sin ^2}2x.\left( {2\cos 2x + 1} \right) = 0$ $ \Leftrightarrow \left[ \begin{array}{l}\sin 2x = 0\\\cos 2x = \frac{1}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}2x = k\pi \\2x = \pm \frac{{2\pi }}{3} + k2\pi \end{array} \right.$$ \Leftrightarrow \left[ \begin{array}{l}x = \frac{{k\pi }}{2}\\x = \pm \frac{\pi }{3} + k\pi \end{array} \right.$$ \Leftrightarrow \left[ \begin{array}{l}x = \frac{{k\pi }}{2}\\x = \frac{{k\pi }}{3}\end{array} \right.$.
$\left( {\sin x - \sin 2x} \right)\left( {\sin x + \sin 2x} \right) = {\sin ^2}3x$ $ \Leftrightarrow {\sin ^2}x - {\sin ^2}2x = {\sin ^2}3x$.
$ \Leftrightarrow \frac{{1 - \cos 2x}}{2} - {\sin ^2}2x = \frac{{1 - \cos 6x}}{2}$ $ \Leftrightarrow \cos 6x - \cos 2x - 2{\sin ^2}2x = 0$
$ \Leftrightarrow - 2\cos 4x.\sin 2x - 2{\sin ^2}2x = 0$ $ \Leftrightarrow 2{\sin ^2}2x.\cos 2x + {\sin ^2}2x = 0$.
$ \Leftrightarrow {\sin ^2}2x.\left( {2\cos 2x + 1} \right) = 0$ $ \Leftrightarrow \left[ \begin{array}{l}\sin 2x = 0\\\cos 2x = \frac{1}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}2x = k\pi \\2x = \pm \frac{{2\pi }}{3} + k2\pi \end{array} \right.$$ \Leftrightarrow \left[ \begin{array}{l}x = \frac{{k\pi }}{2}\\x = \pm \frac{\pi }{3} + k\pi \end{array} \right.$$ \Leftrightarrow \left[ \begin{array}{l}x = \frac{{k\pi }}{2}\\x = \frac{{k\pi }}{3}\end{array} \right.$.