已知函数f(x)=根号3sin²x+sinxcosx-根号3/2(x∈R) 在△abc中,若A

f(x)=根号3sin²x+sinxcosx-根号3/2=根号3(1/2-cos(2x)/2)+1/2*sin2x-根号3/2
=1/2*(sin2x-根号(3)cos2x)=根号(1+3)/2*sin(2x-π/3)=sin(2x-π/3)
f(A)=f(B)=1/2
sin(2A-π/3)=sin(2B-π/3)=1/2,A