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

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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

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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

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