Question

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

Answer 1

$2cos^22x+cos2x+cos6x=1\(2cos^22x-1)+(cos2x+cos6x)=0\cos4x+2cos4xcos2x=0\cos4x(1+2cos2x)=0\cos4x=0\4x=frac{pi}{2}+pi n\x=frac{pi}{8}+frac{pi n}{4},; nin Z;\\[tex]1+2cos2x=0\cos2x=-frac{1}{2}\2x=pm(pi-frac{pi}{3})+2pi n\2x=pm frac{2pi}{3}+2pi n\x=frac{pi}{3}+pi n, ; nin Z$