sincos

2015-05-20 4:03 am
rewrite 3tanθ-2sin(90°+θ)=0 in the form asin²θ+bsinθ+c=0

回答 (3)

2015-05-20 5:09 am
✔ 最佳答案
3 tanθ-2 sin(90°+θ)=0==> 3 sinθ / cosθ-2 cosθ=0==> 3 sinθ-2 cos²θ=0==> 3 sinθ-2(1-sin²θ)=0==> 2 sin²θ+3 sinθ-2=0
2015-05-20 11:43 pm
3tanθ-2sin(90°+θ)=0
(3sinθ)/(cosθ)-2cosθ=0
3sinθ -2(cos²θ)=0
3sinθ-2(1-sin²θ)=0
3sinθ-2+2sin²θ=0
2sin²θ+3sinθ-2=0
2015-05-20 5:12 am
3tanθ-2sin(90°+θ)=0
3sinθ/cosθ -2cosθ = 0
3sinθ - 2cos^2(θ) = 0
3sinθ - 2(1-sin^2(θ))=0
2sin^2(θ)+3sinθ-2=0


收錄日期: 2021-04-15 20:13:46
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150519000051KK00054

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