2kπ-π/2≤x+π/3≤2kπ+π/2,k∈Z
2kπ-5π/6≤x≤2kπ+π/6,k∈z
单调递增区间[0,π/6],单调递减区间[π/6,π/2]
f(x)max=f(π/6)=2sin(π/2)=2
f(0)=√3,f(π/2)=1
所以f(x)min=1