f（x）＝√3/2sin2x-cos^2x-1/2
=√3/2sin2x-1/2*(1+cos(2x)) -1/2
=√3/2sin2x-1/2* cos(2x)-1
= sin(2x-π/6)-1
函数的最小正周期是2π/2=π.
2kπ-π/2≤2x-π/6≤2kπ+π/2,k∈Z.
kπ-π/6≤x≤kπ+π/3,k∈Z.
所以函数的单调增区间是[kπ-π/6,kπ+π/3],k∈Z.