Помогите решить cos2x-cos4x-cos6x=0

[latex]\displaystyle cos2x-cos4x-cos6x=0\\\\cos2x-cos6x=2\frac{2x+6x}{2}cos\frac{2x-6x}{2}=2cos4xcos2x;\\\\2cos4xcos2x-cos4x=0\\\\cos4x(2cosx-1)=0\\\\\\1.\; cos4x=0\\\\ 4x=\frac{\pi}{2}+\pi n\\\\x=\frac{\pi}{8}+\frac{\pi n}{4}, \; n\in Z;\\\\\\2.\; 2cosx-1=0\\cosx=\frac{1}{2}\\x=\pm arccos\frac{1}{2}+2\pi n\\x=\pm \frac{\pi}{3}+2\pi n, \; n\in Z.[/latex]