求函数f(x)=cosx-（根号3）sinx在[0,2π]的单调递减区间

解
f(x)=cosx-√3sinx
=2[(1/2)cosx-(√3/2)sinx]
=2cos(x+π/3)
令2kπ≤x+π/3≤2kπ+π得 2kπ-π/3≤x≤2kπ+2π/3
∵x∈[0,2π],k可取得0 得0≤x≤2π/3 ,k取1得5π/3≤x≤2π
∴f(x)的单调减区间为[0,2π/3]∪[5π/3,2π]