Question

срожно нужно решение 

Answer 1

$cos4x=cos^22x-sin^22x$

Уравнение перепишется так:

$4cos^22x-4sin^22x+6sin^22x+5cos2x=0,; ; sin^22x=1-cos^22x\\4cos^22x+2(1-cos^22x)+5cos2x=0\\2cos^22x+5cos2x+2=0\\t=cos2x,; 2t^2+5t+2=0,; D=25-16=9,\\ t_1=-frac{1}{2},; t_2=-2<-1; to ; cos2x=-2; ne; imeet; reshenij\\cos2x=-frac{1}{2},; 2x=pm (pi -frac{pi}{3})+2pi n=pm frac{2pi }{3}+2pi n,; nin Z\\x=pm frac{pi}{3}+pi n,; nin Z\\x_1=-frac{pi}{3},x_2=frac{pi}{3},; x_3=frac{2pi }{3}; in [-frac{pi}{2},frac{2pi }{3}]$