помогите решить, плиииз!! даю 49 баллов

(√3sin2x+cos2x)²=8-4cos(2x+2π/3)
4*(√3/2sin2x+1/2cos2x)²=8-4cos(2x+2π/3)
4*(cosπ/3*cos2x+sinπ/3*sin2x)=
8-cos(π-π/3+2x)
4*cos²(2x-π/3)=8+4cos(2x-π/3)
cos(2x-π/3)=t€[-1;1]
t²-t-2=0
D=1+8=9=3²
t=(1±3)/2
t1=2;t2=-1
cos(2x-π/3)=-1
2x-π/3=π+2πk
2x=4π/3+2πk
x=2π/3+πk;k€Z