f(x)=1/2*sin2x-√3/2*cos2x+1
=sin(2x-π/3)+1
最小正周期T=2π/2=π
令2kπ-π/2≤2x-π/3≤2kπ+π/2,
得：kπ-π/12≤x≤kπ+5π/12
即单调递增区间为[kπ-π/12,kπ+5π/12] (k∈Z)