若1+(sinα)^2=3sinαcosα.求tanα.

1+(sinα)^2=3sinαcosα
将 1=(sinα)^2+(cosα)^2 代入原方程中便得 2(sinα)^2+(cosα)^2=3sinαcosα
两边同时除以(cosα)^2得
2(tanα)^2-3tanα+1=0
=> (2tanα-1)(tanα-1)=0
tanα=1/2 或 tanα=1

1+(sinα)^2=3sinαcosα
两边除以（cosα）^2

1/（cosα）^2 +(tanα)^2=3tanα
（sec α）^2 +(tanα)^2-3tanα=0
1+（tan α）^2 +(tanα)^2-3tanα=0
1+2(tanα)^2-3tanα=0
(tanα-1)(2tanα-1)=0
tanα=1或者1/2

tanα=1/2或tana=1