如题所述
两边同时除以(cosa)^2,得:2+3tana-3(tana)^2=(seca)^2
2+3tana-3(tana)^2=(tana)^2+1
4(tana)^2-3tana-1=0
(4tana+1)(tana-1)=0
得:tana=-1/4, 或tana=1
2+3tana-3(tana)^2=(tana)^2+1
4(tana)^2-3tana-1=0
(4tana+1)(tana-1)=0
得:tana=-1/4, 或tana=1
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第1个回答 2014-12-13
(1)2cos^2α+3cosαsinα-3sin^2α=1
2+3tanα-3tan^2α=sec^2α
3tan^2α-3tanα-2+(tan^2α+1)=0
4tan^2α-3tanα-1=0
(4tanα+1)(tanα-1)=0
4tanα+1=0
tanα=-1/4
或者tanα-1=0
tanα=1
(2)=(2tanα-3)/(4tanα-9)
当tanα=-1/4时
=[2(-1/4)-3]/[4(-1/4)-9]
=(-7/2)/(-10)
=7/20
当tanα=1时
=(2-3)/(4-9)
=1/5追问
2+3tanα-3tan^2α=sec^2α
3tan^2α-3tanα-2+(tan^2α+1)=0
4tan^2α-3tanα-1=0
(4tanα+1)(tanα-1)=0
4tanα+1=0
tanα=-1/4
或者tanα-1=0
tanα=1
(2)=(2tanα-3)/(4tanα-9)
当tanα=-1/4时
=[2(-1/4)-3]/[4(-1/4)-9]
=(-7/2)/(-10)
=7/20
当tanα=1时
=(2-3)/(4-9)
=1/5追问
之前有个人已经先回答啦,不过还是谢谢咯!
第2个回答 2014-12-13
太难追问
为完成任务要不要那么水
追答哈