Question

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

Answer 1

$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}, ; nin Z;\\1+cos4x=0\cos4x=-1\4x=pi +2pi n\x=frac{pi}{4}+frac{pi n}{2}, ; nin Z.$