f(x)＝√3sinx+cos(π/3+x)
＝√3sinx+cos(π/3)cosx-sin(π/3)sinx
＝√3sinx+1/2cosx-(√3/2)sinx
＝(√3/2)sinx+(1/2)cosx
＝sin(x+π/6).
x+π/6＝2kπ+π/2→x＝2kπ+π/3时,
所求最大值为：1；
x+π/6＝2kπ+3π/2→x＝2kπ+8π/6时,
所求最小值为：-1.