已知函数f(x)=sin(2ωx+π/6)(ω>0)直线x=x1/x=x2是y=f(x)图像的恣意两条对称轴,且丨x1-x2丨最小值为π/2 t已知函数f(x)=sin(2ωx+π/6)(ω>0)直线x=x1/x=x2是y=f(x)图像的恣意两条对称轴,且丨x1-x2丨最小值为π/2(1) 求函数f(x)的单调增区间:(2) 若f(x)=1/3,α∈[-π/3,π/6],求f(α=π/6)的值; (3)若关于x的方程f(x+π/6)+mcosx+3=0在x∈（0,π/2）有实数解,务实数m的取值范围.求协助啊求协助

由丨x1-x2丨最小值为π/2得：T/2=π/2，所以T=π，因此ω=1,即f(x)=sin(2x+π/6)
(1)由-π/2+2kπ≤2x+π/6≤π/2+2kπ得：-π/3+kπ≤x≤π/6+kπ，即函数f(x)的单调增区间为
[-π/3+kπ,π/6+kπ](k∈z)
(2)因为f(x)=1/3,α∈[-π/3,π/6],所以f(α=π/6)=1/3
(3)f(x+π/6)+mcosx+3=0即sin(2x+π/2)+mcosx+3=0，所以cos2x+mcosx+3=0,即
2cos^2x+mcosx+2=0,
因为方程f(x+π/6)+mcosx+3=0在x∈（0，π/2）有实数解，所以有
△≥0
x1+x2>0
解这个不等式组可得：m≤-4

由丨x1-x2丨最小值为π/2得：T/2=π/2,所以T=π,因此ω=1,即f(x)=sin(2x+π/6)
(1)由-π/2+2kπ≤2x+π/6≤π/2+2kπ得：-π/3+kπ≤x≤π/6+kπ,即函数f(x)的单调增区间为
[-π/3+kπ,π/6+kπ](k∈z)
(2)因为f(x)=1/3,α∈[-π/3,π/6],所以f(α=π/6)=1/3
(3)f(x+π/6)+mcosx+3=0即sin(2x+π/2)+mcosx+3=0,所以cos2x+mcosx+3=0,即
2cos^2x+mcosx+2=0,
因为方程f(x+π/6)+mcosx+3=0在x∈（0,π/2）有实数解,所以有
△≥0
x1+x2>0
解这个不等式组可得：m≤-4