解
f(x)=√3sinxcosx-sin^2x+1/2
=√3/2sin2x+1/2(1-2sin^2x)
=√3/2sin2x+1/2cos2x
=sin2xcosπ/6+cos2xsinπ/6
=sin(2x+π/6)
∴最小正周期为：T=2π/2=π
当2x+π/6=π/2+2kπ
即x=π/6+kπ时
f(x)取得最大值为：1
∴取得最大值的x的集合为：{x|x=π/6+kπ,k∈z}不用谢