Giải phương trình $4\left( {{{\sin }^6}x + {{\cos }^6}x} \right) + 2\left( {{{\sin }^4}x + {{\cos }^4}x} \right) = 8 - 4{\cos ^2}2x$
A. $x = \pm \frac{\pi }{3} + \frac{{k\pi }}{2}$, $k \in \mathbb{Z}$.
B. $x = \pm \frac{\pi }{{24}} + \frac{{k\pi }}{2}$, $k \in \mathbb{Z}$.
C. $x = \pm \frac{\pi }{{12}} + \frac{{k\pi }}{2}$, $k \in \mathbb{Z}$.
D. $x = \pm \frac{\pi }{6} + \frac{{k\pi }}{2}$, $k \in \mathbb{Z}$.
A. $x = \pm \frac{\pi }{3} + \frac{{k\pi }}{2}$, $k \in \mathbb{Z}$.
B. $x = \pm \frac{\pi }{{24}} + \frac{{k\pi }}{2}$, $k \in \mathbb{Z}$.
C. $x = \pm \frac{\pi }{{12}} + \frac{{k\pi }}{2}$, $k \in \mathbb{Z}$.
D. $x = \pm \frac{\pi }{6} + \frac{{k\pi }}{2}$, $k \in \mathbb{Z}$.
Chọn C.
Ta có: $4\left( {{{\sin }^6}x + {{\cos }^6}x} \right) + 2\left( {{{\sin }^4}x + {{\cos }^4}x} \right) = 8 - 4{\cos ^2}2x$
$ \Leftrightarrow 4\left( {1 - 3{{\sin }^2}x{{\cos }^2}x} \right) + 2\left( {1 - 2{{\sin }^2}x{{\cos }^2}x} \right) = 8 - 4{\cos ^2}2x$
$ \Leftrightarrow 6 - 4{\sin ^2}2x = 8 - 4{\cos ^2}2x$
$ \Leftrightarrow \cos 4x = \frac{1}{2}$
Ta có: $4\left( {{{\sin }^6}x + {{\cos }^6}x} \right) + 2\left( {{{\sin }^4}x + {{\cos }^4}x} \right) = 8 - 4{\cos ^2}2x$
$ \Leftrightarrow 4\left( {1 - 3{{\sin }^2}x{{\cos }^2}x} \right) + 2\left( {1 - 2{{\sin }^2}x{{\cos }^2}x} \right) = 8 - 4{\cos ^2}2x$
$ \Leftrightarrow 6 - 4{\sin ^2}2x = 8 - 4{\cos ^2}2x$
$ \Leftrightarrow \cos 4x = \frac{1}{2}$
