$\sin^6x+\cos^6x$
$=(\sin^2x+\cos^2x)(\sin^4x+\cos^4x-\sin^2x\cos^2x)$
$=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x-\sin^2x\cos^2x$
$=1-3\sin^2x\cos^2x$
$=1-\dfrac{3}{4}\sin^22x$
Phương trình trở thành:
$1-\dfrac{3}{4}\sin^22x=\sin2x$
$\to \dfrac{3}{4}\sin^22x+\sin2x-1=0$
$\to \left[ \begin{array}{l}\sin2x=\dfrac{2}{3}\\\sin2x=-2(\text{loại})\end{array} \right.$
$\to \left[ \begin{array}{l}2x=\arcsin\dfrac{2}{3}+k2\pi\\2x=\pi-\arcsin\dfrac{2}{3}+k2\pi\end{array} \right.$
$\to \left[ \begin{array}{l}x=\dfrac{1}{2}\arcsin\dfrac{2}{3}+k\pi\\x=\dfrac{\pi}{2}-\dfrac{1}{2}\arcsin\dfrac{2}{3}+k\pi\end{array} \right.$
