Một nghiệm của phương trình ${\cos ^2}x + {\cos ^2}2x + {\cos ^2}3x = 1$ có nghiệm là
A. $x = \frac{\pi }{8}$.
B. $x = \frac{\pi }{{12}}$.
C. $x = \frac{\pi }{3}$.
D. $x = \frac{\pi }{6}$.
A. $x = \frac{\pi }{8}$.
B. $x = \frac{\pi }{{12}}$.
C. $x = \frac{\pi }{3}$.
D. $x = \frac{\pi }{6}$.
Chọn D
${\cos ^2}x + {\cos ^2}2x + {\cos ^2}3x = 1$$ \Leftrightarrow \frac{{1 + \cos 2x}}{2} + \frac{{1 + \cos 4x}}{2} + \frac{{1 + \cos 6x}}{2} = 1$
$ \Leftrightarrow \cos 6x + \cos 2x + 1 + \cos 4x = 0$$ \Leftrightarrow 2\cos 4x\cos 2x + 2{\cos ^2}2x = 0$
$ \Leftrightarrow \left[ \begin{array}{l}\cos 2x = 0\\\cos 4x = \cos \left( {\pi - 2x} \right)\end{array} \right.$$ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{4} + k\pi \\x = \frac{\pi }{6} + k\frac{\pi }{3}\\x = - \frac{\pi }{2} + k\pi \end{array} \right.$, ($k \in \mathbb{Z}$).
${\cos ^2}x + {\cos ^2}2x + {\cos ^2}3x = 1$$ \Leftrightarrow \frac{{1 + \cos 2x}}{2} + \frac{{1 + \cos 4x}}{2} + \frac{{1 + \cos 6x}}{2} = 1$
$ \Leftrightarrow \cos 6x + \cos 2x + 1 + \cos 4x = 0$$ \Leftrightarrow 2\cos 4x\cos 2x + 2{\cos ^2}2x = 0$
$ \Leftrightarrow \left[ \begin{array}{l}\cos 2x = 0\\\cos 4x = \cos \left( {\pi - 2x} \right)\end{array} \right.$$ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{4} + k\pi \\x = \frac{\pi }{6} + k\frac{\pi }{3}\\x = - \frac{\pi }{2} + k\pi \end{array} \right.$, ($k \in \mathbb{Z}$).