解：最小正周期T=2π÷2=π
解由函数f(x)=2sin(ωx+6/π)(ω＞0，x∈R)的最小正周期为π 知T=2π/ω=π 解得ω=2 故f(x)=2sin(2x+π/6) 由f(α)=2/3 知2sin(2a+π/6)=2/3 即sin(2a+π/6)=1/3 α∈(0,π/8) 知2a+π/6是锐角故cos(2a+π/6)=2√2/3 故cos2a =cos(2a+π/6-π/6) =cos(2a+π/6)sinπ/6-sin(2a+π/6)cosπ/6 =2√2/3×1/2-1/3×√3/2 =(2√2-√3)/6
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