f(x)=sin(2x+π/3)-cos(2x+π/6)+2cos²x
f (π/8)=sin7π/12-cos5π/12+2cos²π/8
=sin5π/12-cos5π/12+1+cosπ/4
=√2sin(5π/12-π/4)+1+√2/2
=√2sinπ/6+1+√2/2
=√2/2+1+√2/2
=√2+1 （根号2+1）
答：f(π/8)的值为√2+1,即根号2+1