求函数f(x)=sin(x+π/3)+根号3sin(x-π/6)的单调区间

增区间[2kπ－7π/6,2kπ－π/6] (k∈Z),减区间[2kπ－π/6,2kπ＋5π/6] (k∈Z),

y＝cosx＋cos(x＋π/3)＝cosx＋cosxcosπ/3－sinxsinπ/3＝(3/2)cosx－(√3/2)sinx＝√3[(√3/2)cosx－(1/2)sinx]＝√3[sinπ/3cosx－cosπ/3sinx]＝√3sin(π/3－x)
∴函数y=cosx+cos(x+π/3)的最大值为√3