f(x)=cos(2x+π/3)+cos(2x-π/3)+[1+cos(2x)]/2-1/2
=2cos(2x)cos(π/3)+cos(2x)/2
=[3cos(2x)]/2
所以1,最小正周期是π
2,当x∈[π/8,5π/8]时
2x∈[π/4,5π/4]
所以f(x)∈[-3/2,3√2/4]
f(x)=-3/2时,2x=π
x=π/2
f(x)=3√2/4时,2x=π/4
x=π/8