f（x）=2cos²x-1+√3sin2x+1=cos2x+√3sin2x+1=2（1/2cos2x+√3/2sin2x）+1
=2sin（2x+π/6）+1,所以周期T=2π/2=π.
-π/2+2πn≤2x+π/6≤π/2+2πn,-π/3+πn≤x≤π/6+πn,即递增区间[-π/3+πn,π/6+πn],递减区间[π/6+πn,2π/3+πn]
（2） -π/6＜x＜π/6,则-π/6＜2x+π/6＜π/2
f（x）=2sin（2x+π/6）+1=5/3,sin（2x+π/6）=1/3.cos（2x+π/6）=2√2/3则sin2x=sin（2x+π/6-π/6）=sin（2x+π/6）cosπ/6+sinπ/6cos（2x+π/6）=（√3-2√2）/6