已知函数fx=sin2分之x×cos2分之x+根号3sin²2分之x+2分之根号3.求fx最小

f(x)=sin(x/2)cos(x/2)+√3*sin²(x/2)+√3/2
=1/2*sinx+√3/2*(1-cosx)+√3/2
=1/2*sinx-√3/2*cosx+√3
=sin(x-π/3)+√3
∴f(x)的最小正周期T=2π
令-π/2+2kπ≤x-π/3≤π/2+2kπ
得：-π/6+2kπ≤x≤5π/6+2kπ
∴f(x)的单调递增区间为[-π/6+2kπ,5π/6+2kπ] (k∈Z)