已知函数f(x)＝sinx＋sin(x－3分之π).求f(x)的单调递增区间

题目

已知函数f(x)＝sinx＋sin(x－3分之π).求f(x)的单调递增区间
好纠结

答案

f(x)＝sinx＋sin(x－π/3).
=sinx+sinxcosπ/3-cosxsinπ/3
=3/2*sinx-√3/2*cosx
=√3(sinxcosπ/6-cosxsinπ/6)
=√3sin(x-π/6)
由2kπ-π/2≤ x-π/6≤2kπ+π/2,k∈Z
得：2kπ-π/3≤ x-π/6≤2kπ+2π/3,k∈Z
∴f(x)的单调递增区间是[2kπ-π/3,2kπ+2π/3],k∈Z