答：
y=2√3sinxcosx+2cos²x+1
=√3sin2x+cos2x+1+1
=2*[(√3/2)sin2x+(1/2)cos2x]+2
=2sin(2x+π/6)+2
所以：
最小正周期T=2π/2=π
单调递增区间满足：2kπ-π/2<=2x+π/6<=2kπ+π/2,[kπ-π/3,kπ+π/6]
单调递减区间满足：2kπ+π/2<=2x+π/6<=2kπ+3π/2,[kπ+π/6,kπ+2π/3]
k∈Z