Question

设函数f(x)=(sinwx+coswx)^2+2cos^2wx(w>0)的最小正周期为2π/3

求w的最小正周期

Answer 1

解：f(x)=(sinwx+coswx)^2+2cos^2wx
=1+2sinwxcoswx+(1+cos2wx)
=sin2wx+cos2wx+2
=√2sin(2wx+π/4)+2
T=2π/2w=2π/3
vv=3/2