已知函数f(x)=sin(x/2)+根号3cos(x/2),x属于R,求f(x)最小正周期,在[-2π,2π]上的单调递

f(x)=sin(x/2)+根号3cos(x/2)
=2[1/2* sin(x/2)+根号3/2*cos(x/2)]
=2 sin(x/2+π/3)
最小正周期为4π.
2kπ-π/2≤x/2+π/3≤2kπ+π/2,k∈Z.
4kπ-5π/3≤x≤4kπ+π/3,k∈Z.
因为x∈[-2π,2π],
当k=0时,-5π/3≤x≤π/3.
所以函数在[-2π,2π]上的单调递增区间是[-5π/3,π/3].