f(x)=sin(2x+pi/3)+sin(2x-pi/3)+2cosx^2-1
=2sin2xcos(π/3)+2cosx^2-1
=sin2x+cos2x
=√2sin(2x+π/4)
故最小正周期=2π/2=π

x∈[-π/4,π/4]
2x∈[-π/2,π/2]
2x+π/4∈[-π/4,3π/4]
因此最大值2x+π/4=π/2,为√2
2x+π/4=-π/4,为-1