\(\Leftrightarrow\frac{cos2x.cos4x.cos8x}{cos^2x.cos^22x.cos^24x}=8\)
\(\Leftrightarrow\frac{cos8x}{cos^2x.cos2x.cos4x}=8\)
\(\Leftrightarrow cos8x=8cos^2x.cos2x.cos4x\)
Do \(sinx=0\) ko phải nghiệm
\(\Leftrightarrow sinx.cos8x=cosx.8sinx.cosx.cos2x.cos4x\)
\(\Leftrightarrow sinx.cos8x=cosx.sin8x\)
\(\Leftrightarrow sin\left(8x-x\right)=0\Leftrightarrow sin7x=0\)