Срешением ! (sin2x+√3cos2x)^2-5=cos(п/6-2x)

Sin2x+√3cos2x=2(1/2*sin2x+√3/2*cos2x)=2cos(2x-π/6)
(sin2x+√3cos2x)²=cos(π/6-2x)
4cos²(2x-π/6)-cos(2x-π/6)=0
cos(2x-π/6)*(4cos(2x-π/6)-1)=0
cos(2x-π/6)=0
2x-π/6=π/2+πn,n∈z
2x=2π/3+πn,n∈z
x=π/3+πn/2,n∈z
cos(2x-π/6)=1/4
2x-π/6=+-arccos0,25+2πk,k∈z
2x=+-arccos0,25+π/6+2πk,k∈z
x=+-0,5arccos0,25+π/12+πk,k∈z