已知角α∈(0,π),向量m=(2,cosα),n=(cosα²,1),且mn=1,f（x）=根号3sinx+cosx

mn=1,2cos^2a+cosa=12cos^2a+cosa-1=0(2cosa-1)(cosa+1)=0cosa=1/2,cosa=-1(舍去,因α∈(0,π),)a=π/3f（x）=√3sinx+cosx＝2（√3/2sinx+1/2cosx)=2sin(x+π/6)f（x+α）=2sin(x+π/3+π/6)=2sin(x+π/2)=2cosx故...