Question

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？

Answer 1

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/历此w=π
w=2
∴函数为f(x)=-sin2x
∴函数的单调增区间为：（孙烂桥π/2+kπ,3π/2+kπ）k∈Z
函数的单调减区间为：（-π/2+kπ,π/2+kπ) k∈Z

有不明白的地方再问哟，祝你学习进步则猛，更上一层楼！ (*^__^*)

Answer 2

f(x)=2sinwx-----------w=2
所以与的sinx单调区指亮间一样,在(0,kT/册逗轿2)上州肆增,(kt/2,3kt/2）上减