f(x)=(√3/2)sin2wx－(1/2)cos2wx－(1/2)=sin(2wx－π/6)－(1/2).周期T=2π/|w|=π,则w=1；此时f(x)=sin(2x－π/6)－(1/2)增区间是2kπ－π/2≤2x－π/6≤2kπ＋π/2,得：kπ－π/6≤x≤kπ＋π/3