Question

5cos2x-6cos^2x+4=0 Помогите

Answer 1

5cos2x -6cos²x +4 =0;

5cos2x -6(1+cos2x)/2 +4 =0  ;

cos2x  = -1/2 ;

2x = (+/-) (π -π/3) +2π*k , k∈Z.

x = (+/-) π/3 + π*k , k∈Z.