Question

已知函数f(x)=(根号3sinwx+coswx)coswx-1/2(w>0)的最小正周期为4π.1.求w的值

Answer 1

f(x)=2(√3/2 sinwx+1/2 coswx)coswx-1/2
=2sin (wx+π/6)coswx-1/2
=sin(wx+π/6+wx)+sin(wx+π/6-wx)-1/2
=sin(2wx+π/6)
∴T=2π/2w=4π
即w=1/4