Lời giải:
Ta có:
$\cos (4x+\frac{\pi}{3})=\sin (x+\frac{\pi}{5})=\cos [\frac{\pi}{2}-(x+\frac{\pi}{5})]=\cos (\frac{3}{10}\pi-x)$
$\Leftrightarrow \cos (4x+\frac{\pi}{3})-\cos (\frac{3}{10}\pi-x)=0$
$\Leftrightarrow -2\sin (\frac{3}{2}x+\frac{19}{60}\pi)\sin (\frac{5}{2}x-\frac{\pi}{60})=0$
$\Rightarrow \sin (\frac{3}{2}x+\frac{19}{60}\pi)=0$ hoặc $\sin (\frac{5}{2}x-\frac{\pi}{60})=0$
$\Rightarrow x=\frac{2}{3}k\pi-\frac{19}{90}\pi$ hoặc $x=\frac{2}{5}k\pi+\frac{\pi}{150}$ với $k$ nguyên.