已知函数f(x)=2sin²x+2庚号下3sinxcosx+1.求f(x)的单调递增区间.

已知函数f(x)=2sin²x+2庚号下3sinxcosx+1.求f(x)的单调递增区间.

f(x)=2sin²x+2庚号下3sinxcosx+1
=1-cos2x+√3sin2x+1=2+2sin(2x-π/6)
根据sinx的单调区间可得2kπ-π/2=