(1)cosα+2sinα=－（根号5）
1+4sinα*cosα+3sinα^2=5
4sinαcosα+3sinα^2=4(sinα^2+cosα^2)
4tanα+3tanα^2=4tanα^2+4
tanα=2
(2)tanα=sinα/cosα=1/2
=> cosα=2sinα (1)
∵ sin²α+cos²α=1 (2)
(1)代入(2)得：5sin²α=1
sinα=±√5/5
代入(1)得：cosα=±2√5/5
2sin²α+3sinαcosα
=(2sin²α+3sinαcosα)/(sin²α+cos²α)
=(2tan²α+3tanα)/(tan²α+1)
=(1/2+3/2)/(1/4+1)
=8/5
(3)第三题看不明白