f(x)=cosx/2-√3sinx/2若x∈[-2π,2π],求函数f(x)的单调减区间

题目

f(x)=cosx/2-√3sinx/2若x∈[-2π,2π],求函数f(x)的单调减区间

答案

f(x)=cosx/2-√3sinx/2
=(1/2)cosx--(√3/2)sinx
=cosπ/3cosx-sinπ/3sinx
=cos(x+π/3)
2kπ≤x+π/3≤2kπ+π
2kπ-π/3≤x≤2kπ+2π/3
k=-1时,[-2π,-4π/3]
k=0时,[-π/3,2π/3]
k=1时,[5π/3,2π]
所以减区间有三个[-2π,-4π/3],[-π/3,2π/3],[5π/3,2π]