设函数f(x)(x∈R)满足f(x+π)=f(x)+sinx,0≤x<π时,f(x)=0,则f(11π6)=1212

2025-06-27 14:35:04
推荐回答(1个)
回答1:

函数f(x)(x∈R)满足f(x+π)=f(x)+sinx,0≤x<π时,f(x)=0,
f(

11π
6
)=f(
6
+π)
=f(
6
)+sin
6
=
1
2

故答案为:
1
2