∵与直线y=1的交点中距离最近的两点间距离为π/3
∴f(kπ+π/2)-f(kπ+π/2-π/6)=1
即2sin[ω(kπ+π/2)+φ]-2sin[ω(kπ+π/2-π/6)+φ]=1
且f(kπ+π/2)=2
f(kπ+π/2-π/6)=1
解得ω=1答案是2解:ωx1+φ=π/6+2kπ ①ωx2+φ=5π/6 +2kπ②(k∈Z)②-①，得：ω(x2-x1)=2π/3∵x2-x1=π/3ω=2，