应该是tan²α-(根号3+1)tanα+根号3=0吧令tanα=xx^2-(√3+1)x+√3=0. (x-√3)(x-1)=0, x-√3=0,x1=√3；x-1=0, x2=1,∴x1=tanα=√3, α1=60°x2=tanα=1. α2=45° ∴α=60°或α=45°.望采纳...