f(x)=sin(x+[π/6])+sin(x-[π/6])+cosx+a
=
3
2sinx+[1/2]cosx+
3
2sinx-[1/2]cosx+cosx+a
=
3sinx+cosx+a
=2sin(x+[π/6])+a,
∵x∈[-[π/2],[π/2]],
∴x+[π/6]∈[-[π/3],[2π/3]],
∴-
3≤2sin(x+[π/6])≤2,
∴f(x)在区间[-[π/2],[π/2]]上的最大值为2+a=2,
∴a=0.
故答案为:0
f(x)=sin(x+[π/6])+sin(x-[π/6])+cosx+a
=
3
2sinx+[1/2]cosx+
3
2sinx-[1/2]cosx+cosx+a
=
3sinx+cosx+a
=2sin(x+[π/6])+a,
∵x∈[-[π/2],[π/2]],
∴x+[π/6]∈[-[π/3],[2π/3]],
∴-
3≤2sin(x+[π/6])≤2,
∴f(x)在区间[-[π/2],[π/2]]上的最大值为2+a=2,
∴a=0.
故答案为:0