题目内容
已知α∈(0,
),且2sin2α-sinαcosα-3cos2α=0,则
=______.
| π |
| 2 |
sin(α+
| ||
| sin2α+cos2α+1 |
α∈(0,
),且2sin2α-sinαcosα-3cos2α=0,
所以2tan2α-tanα-3=0,解得tanα=
,tanα=-
(舍去)
cosα=
=
=
=
=
=
=
.
故答案为:
.
| π |
| 2 |
所以2tan2α-tanα-3=0,解得tanα=
| 3 |
| 2 |
| 1 |
| 2 |
cosα=
|
|
|
sin(α+
| ||
| sin2α+cos2α+1 |
| ||||
| 2sinαcosα+2cos2α |
| ||
| 4cosα |
| ||||
4×
|
| ||
| 8 |
故答案为:
| ||
| 8 |
练习册系列答案
相关题目