设函数f(x)=(sinwx+coswx)²+2cos²wx(w>0)的最小正周期为2π/3,求：

f(x)=(sinwx+coswx)²+2cos²wx
=[(根号2)*cos(wx-π/4)]²+2cos²wx
=2*[cos²(wx-π/4)+cos²wx]
=cos(2wx-π/2)+cos2wx+2
=sin2wx+cos2wx+2
=(根号2)sin(2wx+π/4)+2
(1)最小正周期T=2π/2w=2π/3,w=3/2
(2)g(x)=(根号2)sin[3(x-π/2)+π/4]+2
=(根号2)sin(3x-5π/4)+2
2kπ-π/2≤3x-5π/4≤2kπ+π/2
解得y=g(x)的单调增区间为[2kπ/3+π/4,2kπ/3+7π/12],k取整数的每一个闭区间