$\sin4x-\sin6x=\sqrt3(\cos6x+\cos4x)$
$\to \sin4x-\sqrt3\cos4x=\sin6x+\sqrt3\cos6x$
$\to \dfrac{1}{2}\sin4x-\dfrac{\sqrt3}{2}\cos4x=\dfrac{1}{2}\sin6x+\dfrac{\sqrt3}{2}\cos6x$
$\to \sin\left(4x-\dfrac{\pi}{3}\right)=\sin\left( 6x+\dfrac{\pi}{3}\right)$
$\to \left[ \begin{array}{l}6x+\dfrac{\pi}{3}=4x-\dfrac{\pi}{3}+k2\pi \\6x+\dfrac{\pi}{3}=\pi-4x+\dfrac{\pi}{3}+k2\pi \end{array} \right.$
$\to \left[ \begin{array}{l}x=\dfrac{-\pi}{3}+k\pi \\x=\dfrac{\pi}{10}+\dfrac{k\pi}{5} \end{array} \right.$
