Giải thích các bước giải:
$\cos 5x\cos x=\cos 4x\cos 2x+3\cos^2x+1$
$\to \dfrac{1}{2}(\cos 6x+\cos 4x)=\dfrac{1}{2}(\cos 6x+\cos 2x)+3\cos^2x+1$
$\to \cos 6x+\cos 4x=\cos 6x+\cos 2x+6\cos^2x+2$
$\to 2\cos ^22x-1=\cos 2x+6.\dfrac{1+\cos 2x}{2}+2$
$\to 2\cos ^22x-1=\cos 2x+3(1+\cos 2x)+2$
$\to \cos ^22x=2\cos 2x+3$
$\to \cos ^22x-2\cos 2x-3=0$
$\to (\cos 2x+1)(\cos 2x-3)=0$
$\to \cos 2x+1=0$
$\to \cos 2x=-1$
$\to x=\dfrac{\pi }{2}+k\pi $