函数y=cos(2x+φ)(-π≤φ<π)的图象向右平移[π/2]个单位后,与函数y=sin(2x+π3)的图象重合,则
1个回答

∵f(x)=cos(2x+φ)=sin[[π/2]+(2x+φ)]=sin(2x+[π/2]+φ),

∴f(x-[π/2])=sin[2(x-[π/2])+[π/2]+φ)]=sin(2x-[π/2]+φ),

又f(x-[π/2])=sin(2x+[π/3]),

∴sin(2x-[π/2]+φ)=sin(2x+[π/3]),

∴φ-[π/2]=2kπ+[π/3],

∴φ=2kπ+[5π/6],又-π≤φ<π,

∴φ=[5π/6].

故选:A.