Giải phương trình $8\cot 2x = \frac{{\left( {{{\cos }^2}x - {{\sin }^2}x} \right).\sin 2x}}{{{{\cos }^6}x + {{\sin }^6}x}}$.
A. $x = - \frac{\pi }{4} + k\pi $.
B. $x = \pm \frac{\pi }{4} + \frac{{k\pi }}{2}$.
C. $x = \frac{\pi }{4} + k\pi $.
D. $x = \frac{\pi }{4} + \frac{{k\pi }}{2}$.
A. $x = - \frac{\pi }{4} + k\pi $.
B. $x = \pm \frac{\pi }{4} + \frac{{k\pi }}{2}$.
C. $x = \frac{\pi }{4} + k\pi $.
D. $x = \frac{\pi }{4} + \frac{{k\pi }}{2}$.
Chọn D.
Điệu kiện: $\left\{ \begin{array}{l}\sin 2x \ne 0\\{\cos ^6}x + {\sin ^6}x \ne 0\end{array} \right. \Leftrightarrow x \ne k\frac{\pi }{2}$
${\rm{pt}} \Leftrightarrow 8\frac{{\cos 2x}}{{\sin 2x}} = \frac{{\cos 2x.\sin 2x}}{{1 - 3{{\sin }^2}x{{\cos }^2}x}} \Leftrightarrow 8\cos 2x\left( {1 - 3{{\sin }^2}x{{\cos }^2}x} \right) = \cos 2x{\sin ^2}2x$
$ \Leftrightarrow \cos 2x\left( {8 - 6{{\sin }^2}2x - {{\sin }^2}2x} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}\cos 2x = 0\\{\sin ^2}2x = \frac{8}{7}\left( {VN} \right)\end{array} \right. \Leftrightarrow x = \frac{\pi }{4} + k\frac{\pi }{2}$.
Điệu kiện: $\left\{ \begin{array}{l}\sin 2x \ne 0\\{\cos ^6}x + {\sin ^6}x \ne 0\end{array} \right. \Leftrightarrow x \ne k\frac{\pi }{2}$
${\rm{pt}} \Leftrightarrow 8\frac{{\cos 2x}}{{\sin 2x}} = \frac{{\cos 2x.\sin 2x}}{{1 - 3{{\sin }^2}x{{\cos }^2}x}} \Leftrightarrow 8\cos 2x\left( {1 - 3{{\sin }^2}x{{\cos }^2}x} \right) = \cos 2x{\sin ^2}2x$
$ \Leftrightarrow \cos 2x\left( {8 - 6{{\sin }^2}2x - {{\sin }^2}2x} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}\cos 2x = 0\\{\sin ^2}2x = \frac{8}{7}\left( {VN} \right)\end{array} \right. \Leftrightarrow x = \frac{\pi }{4} + k\frac{\pi }{2}$.