函数f(x)=sin(x+π/6)+sin(x-π/6)+cosx的最小正周期?

题目

函数f(x)=sin(x+π/6)+sin(x-π/6)+cosx的最小正周期?
函数在【0,2π】上的单调递减区间是?

答案

f(x)=2sinxcos(π/6)+cosx
=√3sinx+cosx
=2sin(x+π/6)
周期为2π
2kπ+π/2≤x+π/6≤2kπ+3π/2,k∈Z.
解得2kπ+π/3≤x≤2kπ+4π/3
在[0,2π]上的单调递减区间是[π/3,4π/3]