Ta có:$VT=2sin8xcosx$
$VF=2[cos(\frac{\pi}{4}-x)+cos(\frac{\pi}{4}+2x)][cos(\frac{\pi}{4}-x)-cos(\frac{\pi}{4}+2x)]$
$=2[2cos(\frac{\pi}{4}+\frac{x}{2})cos(\frac{3x}{2}).2sin(\frac{\pi}{4}+\frac{x}{2})sin(\frac{3x}{2})$
$=2sin3x.sin(\frac{\pi}{2}+x)=2sin3x.cosx$
pt$\Leftrightarrow sin8x.cosx-sin3x.cosx=0$