Đáp án:
$\begin{array}{l}
Dkxd:sin2x \ne 0 \Rightarrow x \ne \dfrac{{k\pi }}{2}\\
{\sin ^6}x + {\cos ^6}x\\
= {\left( {{{\sin }^2}x} \right)^3} + {\left( {{{\cos }^2}x} \right)^3}\\
= \left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^4}x - {{\sin }^2}x.{{\cos }^2}x + {{\cos }^4}x} \right)\\
= {\sin ^4}x + 2{\sin ^2}x.{\cos ^2}x + {\cos ^4}x - 3{\sin ^2}x.{\cos ^2}x\\
= {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - \dfrac{3}{4}{\sin ^2}2x\\
= 1 - \dfrac{3}{4}{\sin ^2}2x\\
Pt:8cot2x.\left( {{{\sin }^6}x + {{\cos }^6}x} \right) = \dfrac{1}{2}\sin 4x\\
\Rightarrow 8.\dfrac{{\sin 2x}}{{\cos 2x}}.\left( {1 - \dfrac{3}{4}{{\sin }^2}2x} \right) = \dfrac{1}{2}.2.\sin 2x.\cos 2x\\
\Rightarrow 8\dfrac{{\sin 2x}}{{\cos 2x}}.\dfrac{{4 - 3{{\sin }^2}2x}}{4} = \sin 2x.\cos 2x\\
\Rightarrow \sin 2x.\left( {\dfrac{{2.\left( {4 - 3{{\sin }^2}2x} \right)}}{{\cos 2x}} - \cos 2x} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sin 2x = 0\left( {ktm} \right)\\
\dfrac{{2\left( {4 - 3{{\sin }^2}2x} \right)}}{{\cos 2x}} = \cos 2x
\end{array} \right.\\
\Rightarrow 8 - 6{\sin ^2}2x = {\cos ^2}2x\\
\Rightarrow 8 - 6{\sin ^2}2x = 1 - {\sin ^2}2x\\
\Rightarrow {\sin ^2}2x = \dfrac{7}{5}\left( {ktm} \right)\\
Do:0 < {\sin ^2}2x \le 1
\end{array}$
Vậy pt vô nghiệm.
