已知函数f(x)=2sin(x-π/3)cosx+sinxcosx+√3sin^2x(x属于R),求f(x)的单调递增区间

题目

已知函数f(x)=2sin(x-π/3)cosx+sinxcosx+√3sin^2x(x属于R),求f(x)的单调递增区间

答案

f(x)=2sin(x-π/3)cosx+sinxcosx+√3sin^2x
=2（sinxcosπ/3-cosxsinπ/3)cosx+sinxcosx+√3sin^2x
=sinxcosx-√3cos^2x+sinxcosx+√3sin^2x
=2sinxcosx-√3(cos^2x-sin^2x)
=sin2x-√3cos2x
=2(1/2sin2x-√3/2cos2x)
=2sin(2x-π/3)
由2kπ-π/2≤2x-π/3≤2kπ+π/2,k∈Z
得kπ-π/12≤x≤kπ+5π/12,k∈Z
∴f(x)递增区间为[kπ-π/12,kπ+5π/12],k∈Z