數學~三角單元~已知2sinθ=-3cotθ 求cosθ?

2sinθ=-3cotθ

2sinθ=-3(cosθ/sinθ)

2(sinθ)^2=-3cosθ

2*[1-(cosθ)^2]+3cosθ=0

-2(cosθ)^2+3cosθ+2=0

2(cosθ)^2-3cosθ-2=0

(cosθ-2)(2cosθ+1)=0

cosθ=2>1(不合)

cosθ= -1/2

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2sinθ= -3cotθ 2 (sinθ)^2= -3cosθ 2(1-(cosθ) ^2)=-3 cosθ令cosθ=c2c^2 -3c -2=0c= -1/2 , 2 (不合)信欣茗
http://tw.myblog.yahoo.com/wang620628/article?mid=447&next=446&l=f&fid=5
好笑的喔

2sinθ = -3cotθ
2sinθ = -3(cosθ/sinθ)
2sin²θ = -3cosθ
2(1 - cos²θ) = -3cosθ
2 - 2cos²θ = -3cosθ
2cos²θ - 3cosθ - 2 = 0
(2cosθ + 1)(cosθ - 2)= 0
cosθ = -1/2 或 cosθ = 2 (捨棄)

所以 cosθ = -1/2