a·b=sin²ωx+√3sinωxsin(ωx+π/2)
=sin²ωx+√3sinωxcosωx
=1/2-1/2cos2ωx+√3/2sin2ωx
=1/2+sin(2ωx-π/3)
f(x)=a·b-2m=sin(2ωx-π/3)-2m+1/2,周期为T=2π/2ω=π/ω
(1)f(x)的相邻的两条对称轴间的距离为半周期,所以T/2=π/2ω=π/2,ω=1,
(2)f(x)=sin(2x-π/3)-2m+1/2的单调增区间[-π/12+kπ,5π/12+kπ];
单调减区间[5π/12+kπ,11π/12+kπ];k∈Z
(3)f(x)在[0,5π/12]上单调递增,在[5π/12,2π/3]单调递减,x=5π/12为一条对称轴,
所以要使f(x)在区间[0,2π/3]上有两个零点只需f(5π/12)＞0,f(2π/3)≤0
联立解得1/4≤m≤3/4