设函数f(x)=sin(wx+2π/3)+sin(wx-2π/3)（w＞0）的最小正周期为π,求函数的单调区间?

f(x)=sin(wx+2π/3)+sin(wx-2π/3)=sinwxcos(2π/3)+coswxsin(2π/3)+sinwxcos(2π/3)-coswxsin(2π/3)=sinwxcos(2π/3)+sinwxcos(2π/3)=2sinwxcos(2π/3)=-sinwx∵T=π∴2π/w=πw=2∴函数为f(x)=-sin2x∴函数的单...