Помогите с решением пожалуйста!!! (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)=0cos(2x-π/6)*(4cos(2x-π/6)-1)=0cos(2x-π/6)=02x-π/6=π/2+πn,n∈z2x=2π/3+πn,n∈zx=π/3+πn/2,n∈zcos(2x-π/6)=1/42x-π/6=+-arccos0,25+2πk,k∈z2x=+-arccos0,25+π/6+2πk,k∈zx=+-0,5arccos0,25+π/12+πk,k∈z