∵0≤x≤π，∴

π

4

≤x+

π

4

≤

5π

4

，∴−

2

2

≤sin(x+

π

4

)≤1，−1≤

2

sin(x+

π

4

)≤

2

．
又∵f（x）=

2

sin(x+

π

4

)＝k在[0，π]上有两解，∴1≤k＜

2

．
∴实数k的取值范围是1≤k＜

2

．
故答案为1≤k＜

2

．