已知函数f（x）=2cos2ωx+2sinωxcosωx+1（x∈R,ω＞0）的最小值正周期是π/2

f(x)=2cos 2wx+2sin wx cos wx +1
=2cos 2wx+sin 2wx +1
=√5 sin (2wx+a) +1 ,a=arctan2
T=π/2,则:2w=4,w=2.
于是：f(x)=√5 sin(4x+a)+1,a=arctan2
y=f(x+φ)=√5 sin(4x+4φ+a)是偶函数,于是：4φ+a=π/2.
于是：φ=π/4-(1/4)arctan2 ,y=√5 sin(4x+π/2)