f(x)=sin2x-(√3/2)cos2x+(1/2)sin2x=√3sin(2x-π/6)
所以最小正周期=π；
当：2kπ-π/2≤2x-π/6≤2kπ+π/2
kπ-π/6≤x≤kπ+π/3
所以增区间为：[kπ-π/6,kπ+π/3]