Lời giải:
$\cos (\frac{\pi}{3}-x)=\sin (2x+\frac{\pi}{3})=\cos [\frac{\pi}{2}-(2x+\frac{\pi}{3})]=\cos (\frac{\pi}{6}-2x)$
$\Leftrightarrow \cos (\frac{\pi}{3}-x)-\cos (\frac{\pi}{6}-2x)=0$
$\Leftrightarrow -2\sin (\frac{\pi}{4}-\frac{3}{2}x)\sin (\frac{\pi}{12}+\frac{x}{2})=0$
$\Rightarrow \sin (\frac{\pi}{4}-\frac{3}{2}x)=0$ hoặc $\sin (\frac{\pi}{12}+\frac{x}{2})=0$
$\Rightarrow x=\frac{\pi}{6}-\frac{2}{3}k\pi$ hoặc $x=2k\pi -\frac{\pi}{6}$ với $k$ nguyên.