f(x)=2cosx^2+√3sin2x+a
=cos2x+1+√3sin2x+a
=2(1/2cos2x+√3/2sin2x)+1+a
=2sin(2x+π/6)+1+a
f(x)在[-π/6,π/6],有：-π/6≤2x+π/6≤π/2
所以可得：当2x+π/6=π/2时有,最大值为 :3+a
最小值为 当2x+π/6=-π/6 时有,最小值为：a
则有：3+a+a=3 所以可得：a=0