已知函数f(X)=sin(x+π\3)-根号3cos(2π\3-x)（1）求函数最小正周期.（2）求函数在区间【-π\2

f(x)＝sin(x＋π/3)－√3cos(2π/3－x)＝sin(x＋π/3)＋√3cos[π－(2π/3－x)]
＝sin(x＋π/3)＋√3cos(x＋π/3)＝2[1/2×sin(x＋π/3)＋√3/2×cos(x＋π/3)]
＝2sin[(x＋π/3)＋π/3]＝2sin[(x＋2π/3)
(1)T＝2π
(2)∵x∈[﹣π/2,π/2] ∴x＋2π/3∈[π/6,7π/6] ∴sin[(x＋2π/3)∈[﹣1/2,1]
∴f(x)∈[﹣1,2]
∴最大值为2,最小值为﹣1