Giải phương trình ${\sin ^2}x + {\sin ^2}3x = {\rm{co}}{{\rm{s}}^2}x + {\rm{co}}{{\rm{s}}^2}3x$
A. $x = \pm \frac{\pi }{4} + k2\pi $, $k \in \mathbb{Z}$.
B. $x = - \frac{\pi }{4} + \frac{{k\pi }}{2},x = \frac{\pi }{8} + \frac{{k\pi }}{4}$, $k \in \mathbb{Z}$.
C. $x = \frac{\pi }{4} + \frac{{k\pi }}{2},x = \frac{\pi }{8} + \frac{{k\pi }}{4}$, $k \in \mathbb{Z}$.
D. $x = - \frac{\pi }{4} + \frac{{k\pi }}{2},x = \frac{\pi }{4} + \frac{{k\pi }}{2}$, $k \in \mathbb{Z}$.
A. $x = \pm \frac{\pi }{4} + k2\pi $, $k \in \mathbb{Z}$.
B. $x = - \frac{\pi }{4} + \frac{{k\pi }}{2},x = \frac{\pi }{8} + \frac{{k\pi }}{4}$, $k \in \mathbb{Z}$.
C. $x = \frac{\pi }{4} + \frac{{k\pi }}{2},x = \frac{\pi }{8} + \frac{{k\pi }}{4}$, $k \in \mathbb{Z}$.
D. $x = - \frac{\pi }{4} + \frac{{k\pi }}{2},x = \frac{\pi }{4} + \frac{{k\pi }}{2}$, $k \in \mathbb{Z}$.
Chọn C.
Phương trình $ \Leftrightarrow {\sin ^2}x - {\cos ^2}x = {\cos ^2}3x - {\sin ^2}3x$
$ \Leftrightarrow \cos 6x + \cos 2x = 0$$ \Leftrightarrow 2\cos 4x.\cos 2x = 0$
$ \Leftrightarrow \left[ \begin{array}{l}\cos 4x = 0\\\cos 2x = 0\end{array} \right.$ $ \Leftrightarrow \left[ \begin{array}{l}4x = \frac{\pi }{2} + k\pi \\2x = \frac{\pi }{2} + k\pi \end{array} \right.$$ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{8} + \frac{{k\pi }}{4}\\x = \frac{\pi }{4} + \frac{{k\pi }}{2}\end{array} \right.,\left( {k \in \mathbb{Z}} \right)$
Phương trình $ \Leftrightarrow {\sin ^2}x - {\cos ^2}x = {\cos ^2}3x - {\sin ^2}3x$
$ \Leftrightarrow \cos 6x + \cos 2x = 0$$ \Leftrightarrow 2\cos 4x.\cos 2x = 0$
$ \Leftrightarrow \left[ \begin{array}{l}\cos 4x = 0\\\cos 2x = 0\end{array} \right.$ $ \Leftrightarrow \left[ \begin{array}{l}4x = \frac{\pi }{2} + k\pi \\2x = \frac{\pi }{2} + k\pi \end{array} \right.$$ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{8} + \frac{{k\pi }}{4}\\x = \frac{\pi }{4} + \frac{{k\pi }}{2}\end{array} \right.,\left( {k \in \mathbb{Z}} \right)$