Lời giải:
PT $\Leftrightarrow (\cos 10x-\cos 6x)+(1-\cos 8x)=0$
$\Leftrightarrow -2\sin 8x\sin 2x+2\sin ^24x=0$
$\Leftrightarrow \sin 8x\sin 2x-\sin ^24x=0$
$\Leftrightarrow 2\sin 4x\cos 4x\sin 2x-\sin ^24x=0$
$\Leftrightarrow \sin 4x[2\cos 4x\sin 2x-\sin 4x]=0$
$\Leftrightarrow \sin 4x[2\cos 4x\sin 2x-2\sin 2x\cos 2x]=0$
$\Leftrightarrow 2\sin 4x\sin 2x(\cos 4x-\cos 2x)=0$
$\Leftrightarrow 2\sin 4x\sin 2x(2\cos ^22x-1-\cos 2x)=0$
$\Leftrightarrow 2\sin 4x\sin 2x(2\cos 2x+1)(\cos 2x-1)=0$
Đến đây thì dễ rồi.
Chắc cái đầu là cos10x?
\(\Leftrightarrow cos10x-cos6x+1-cos8x=0\)
\(\Leftrightarrow-2sin8x.sin2x+2sin^24x=0\)
\(\Leftrightarrow-4sin4x.cos4x.sin2x+2sin^24x=0\)
\(\Leftrightarrow-sin4x.cos4x.sin2x+sin4x.sin2x.cos2x=0\)
\(\Leftrightarrow sin4x.sin2x\left(cos2x-cos4x\right)=0\)
\(\Leftrightarrow sinx.sin2x.sin3x.sin4x=0\)
\(\Leftrightarrow sin3x.sin4x=0\)
\(\Leftrightarrow...\)
