应该是tan²α-(根号3+1)tanα+根号3=0吧
令tanα=x
x^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°.
望采纳哦
应该是tan²α-(根号3+1)tanα+根号3=0吧
令tanα=x
x^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°.
望采纳哦