令x=0,
f(1)=1
f(0)=0
令x=π/2,
f(0)=cos(kπ/2)=0
f(1)=sin(kπ/2)=1
所以kπ/2=2mπ+π/2
kπ=4mπ+π
k=4m+1 (m为整数)
所以k为整数
令x=(2π)/k
f(cos(2π/k))=1
f(sin(2π/k))=0
所以cos(2π/k)=1
sin(2π/k)=0
所以2π/k=2nπ
k=1/n
因为n为整数
所以n=1才能满足k为整数k=1