f(x)=mn=2cos^2x+2√3sinxcosx+a-1+1
=cos2x+√3sin2x+a+1
=2sin(2x+π/6)+a+1
f(x)=0
sin(2x+π/6)=(-a-1)/2
f(x)在【0,π/2】的值域为【-1,2】
f(0)=1
f(x)对称轴：2x+π/6=π/2
x=π/6
π/6*2=π/3＜π/2
f(x)在【π/3,π/2】有唯一解
sin(2x+π/6)在【π/3,π/2】的值域为【-1/2,1/2】
-1/2≤(-a-1)/2＜1/2
-2＜a≤0