f(0)=0⇒f(0)=Asin(ω×0+ϕ)=Asinϕ=0⇒ϕ=kπ,k∈Z⇒f(x)=Asin(ωx+ϕ)(A>0,ω>0,x∈R)是奇函数.f(x)为奇函数⇒ϕ=kπ,k∈Z⇒f(0)=Asin(ω×0+kπ)=Asinkπ=0.所以f(0)=0是f(x)为奇函数的充要条件.
【解析】
若f(0)=0,
则f(0)=Asin(ω×0+ϕ)=Asinϕ=0,
∴ϕ=kπ,k∈Z,
∴f(x)=Asin(ωx+ϕ)(A>0,ω>0,x∈R)是奇函数.
若f(x)为奇函数,
则ϕ=kπ,k∈Z,
∴f(0)=Asin(ω×0+kπ)=Asinkπ=0.
所以f(0)=0是f(x)为奇函数的充要条件.
故答案为:充要.