Решите уравнение: 2cos^2x+cos2x+cos6x=1

[latex]2cos^2x+cos2x+cos6x=1\\(2cos^2x-1)+(cos2x+cos6x))=0\\\\cos2x+cos6x=2cos4xcos2x;\\2cos^2x-1=cos2x;\\\\cos2x+2cos4xcos2x=0\\cos2x(1+2cos4x)=0\\\\cos2x=0\\2x=\frac{\pi}{2}+\pi n\\x=\frac{\pi}{4}+\frac{\pi n}{2}, \; n\in Z;\\\\1+cos4x=0\\cos4x=-1\\4x=\pi +2\pi n\\x=\frac{\pi}{4}+\frac{\pi n}{2}, \; n\in Z.[/latex]