已知函数f(x)=2sin(x-兀/6)cosx+2cos^2x.求(1)f(x)的单调递增区间(2)设兀/4

问题描述:

已知函数f(x)=2sin(x-兀/6)cosx+2cos^2x.求(1)f(x)的单调递增区间(2)设兀/4

(1)f(x)=2sin(x-π/6)cosx+2cos^2x=(√3sinx-cosx)cosx+2cosxcosx
=(√3sinx+cosx)cosx=√3/2sin2x+1/2cos2x+1/2
=sin(2x+π/6)+1/2
令2kπ-π/2