求最小正周期f（x）=sin（2x+π/3）-根号3sin平方x+sinxcosx+（根号3）/2

f(x)=sin(2x+π/3) -√3sin²x+sinxcosx+√3/2
=sin2xcos(π/3)+cos2xsin(π/3)-√3sin²x+sinxcosx+√3/2
=(1/2)sin2x +(√3/2)cos2x-√3sin²x+sinxcosx+√3/2
=sinxcosx+√3cos²x-√3/2-√3sin²x+sinxcosx+√3/2
=2sinxcosx+√3cos2x
=sin2x+√3cos2x
=2[(1/2)sin(2x)+(√3/2)cos(2x)]
=2[sin(2x)cos(π/3)+cos(2x)sin(π/3)]
=2sin(2x+π/3)
T=2π/2=π