∵cos(π+α)=-cosα
∴-cosα=-1/2
cosα=1/2
∴α=π/3或α=-π/3
sin(2π+α)-tan(α-3π)
=sinα+tan(3π-α)
=sinα+tan(2π+π-α)
=sinα+tan(π-α)
=sinα-tanα
則當α=π/3時,
sin(2π+α)-tan(α-3π)
=sinπ/3-tanπ/3
=√3/2-√3
=-√3/2
則當α=-π/3時,
sin(2π+α)-tan(α-3π)
=sin(-π/3)-tan(-π/3)
=-sinπ/3+tanπ/3
=-√3/2+√3
=√3/2