y=sin(x/2)+√3cos(x/2)=2[(1/2)sin(x/2)+(√3/2)cos(x/2)]
=2[cos(π/3)sin(x/2)+sin(π/3)cos(x/2)]
=2sin(x/2+π/3)
1.由于-1≤sin(x/2+π/3)≤1
所以-2≤y≤2
最小正周期T=(2π)/(1/2)=4π
2.要求函数y的单调递增区间,只需求f(x)=sin(x/2+π/3)的单调递增区间,
由于f(x)=sin(x/2+π/3)在[2kπ,2kπ+π/2]上单调递增,所以
2kπ≤x/2+π/3≤2kπ+π/2 (k∈Z)
解得4kπ-2π/3≤x≤4kπ+π/3 (k∈Z)
即函数y=2sin(x/2+π/3)的单调递增区间是[4kπ-2π/3,4kπ+π/3],(k∈Z)