F (x) = 2cos 2x+2 √ 3 SINCOSX+A followed by sin(cosx) or sinx*cosx. According to the latter, if it is the former, it is not a first-grade problem.

f(x)= 2 *(cos2x+ 1)/2+√3 * sin(2x)

=cos2x+√3*sin2x+a+ 1

= 2cos(2x-π/3)+a+ 1；

So T=2π/ω=π.

-π/6≤x≤π/3； -π/3≤2x≤2π/3； -2π/3≤2x-π/3≤π/3；

-1≤ 2cos (2x-π/3) ≤1,so a≤f(x)≤a+2.

So 2a+2 = 3 and a = 1/2.