已知函数f(x)=sin(ωx+π/3),f(π/6)=f（π/4）,且f(x)在区间（π/6,π/4）有最小值无最大值,则ω=

f(π/6)=sin(ωπ/6+π/3),f(π/4)=sin(ωπ/4+π/3),sin(ωπ/6+π/3)=sin(ωπ/4+π/3),ωπ/6+π/3=ωπ/4+π/3,ωπ/6+π/3=π-ωπ/4-π/3,解得：ω=0,ω=4/5,取ω=4/5,f(x)在区间（π/6,π/4）有最大值无最小...答案还有52/5、148/5正弦函数在一、二和三、四象限函数值相等，ωπ/6+π/3＜ωπ/4+π/3，当在一、二象限时，ωπ/6+π/3-2kπ=2kπ+π-(ωπ/4+π/3)，ω=48k/5+4/5，ωπ/6+π/3-2kπ＜π/2，ω＜4k+2，k＜1/4；当在三、四象限时，ωπ/6+π/3-(2kπ+π)=2kπ+2π-(ωπ/4+π/3)，ω=48k/5+28/5，ωπ/6+π/3-(2kπ+π)＜π/2，ω＜12k+7，k＞-7/12；综上k的取值范围：-7/12＜k＜1/4，因为k为整数，所以取k=0，ω=4/5或ω=28/5