应该是
f(x)=sin^2x+√3sinxcosx=-(-2sin^2x\2)+(√3\2)sin2x
=-[(1-2sin^2x)\2]+(√3\2)sin2x+1\2
=-(1\2)cos2x+(√3\2)sin2x+1\2
=sin(2x-π\3)+1\2
所以当2x-π\3=π\2 x=5π\12 时最大值为1+1\2=3\2