已知函数f(x)=2sin(x-π/6)cosx+2cos^2x 求f(x)的单调增区间

问题描述:

已知函数f(x)=2sin(x-π/6)cosx+2cos^2x 求f(x)的单调增区间

y=2sin(x-pi/6)cosx+2(cosx)^2 =2(sinxcospi/6-cosxsinpi/6)cosx+2(cosx)^2 =√3sinxcosx+(cosx)^2 =2sinxcosx*√3/2+(1+cos2x)/2 =√3/2*sin2x+1/2*cos2x+1/2 =sin(2x+pi/6)+1/2 所以A=1,w=2,φ=pi/6,H=1/2 T=2pi/w=pi 所以-pi/2+2kpi