f(x)=cosx+2sinx=sqrt(5)sin(x+arccos(2/sqrt(5)))因x属于[0,π/2]，故x+arccos(2/sqrt(5))属于[arccos(2/sqrt(5)),π/2+arccos(2/sqrt(5))]所以最小值在端点取到（由sinx的增减性）比较f(0)和f(π/2)的大小f(0)=1，f(π/2)=2，所以最小值为1
