∵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.
∵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.