1.rewrite the equation 3tanθ=2cosθ in the form asin²θ+bsinθ+c=0,where a,b and c are integers.Hence solve the equation for 0°<θ<360°
2.Slove sin²θ-3cosθ-1=0 for 0°<θ<360°.
3.(a)Find x if sinx =1/2 and 90°<180°.
(b)Simplify 1-sin²A / cosA
maths
2007-07-24 6:56 pm
回答 (3)
2007-07-24 7:14 pm
✔ 最佳答案
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參考: My Maths knowledge
2007-07-24 8:33 pm
1.3tanθ=2cosθ
3(sinθ/cosθ)=2cosθ
3sinθ=2cos^2θ
3sinθ=2-2sin^2θ
2sin^2θ+3sinθ-2=0
(2sinθ-1)(sinθ+2)=0
sinθ=1/2 OR sinθ= -2(rejected)
θ= 30° OR 150°
2.sin²θ-3cosθ-1=0
1-cos^2θ-3cosθ-1=0
cosθ(cosθ+3)=0
cosθ=0 OR cosθ= -3(rejected)
θ= 90° OR 270°
3A)sinx =1/2
x= 150°
B)1-sin²A / cosA =cos^2A/cosA=cosA
3(sinθ/cosθ)=2cosθ
3sinθ=2cos^2θ
3sinθ=2-2sin^2θ
2sin^2θ+3sinθ-2=0
(2sinθ-1)(sinθ+2)=0
sinθ=1/2 OR sinθ= -2(rejected)
θ= 30° OR 150°
2.sin²θ-3cosθ-1=0
1-cos^2θ-3cosθ-1=0
cosθ(cosθ+3)=0
cosθ=0 OR cosθ= -3(rejected)
θ= 90° OR 270°
3A)sinx =1/2
x= 150°
B)1-sin²A / cosA =cos^2A/cosA=cosA
2007-07-24 7:16 pm
1.3tanθ=2cosθ
3(sinθ/cosθ)=2cosθ
3sinθ=2cos^2θ
3sinθ=2-2sin^2θ
2sin^2θ+3sinθ-2=0
(2sinθ-1)(sinθ+2)=0
sinθ=1/2 OR sinθ= -2(rejected)
θ= 30° OR 150°
2.sin²θ-3cosθ-1=0
1-cos^2θ-3cosθ-1=0
cosθ(cosθ+3)=0
cosθ=0 OR cosθ= -3(rejected)
θ= 90° OR 270°
3A)sinx =1/2
x= 150°
B)1-sin²A / cosA =cos^2A/cosA=cosA
3(sinθ/cosθ)=2cosθ
3sinθ=2cos^2θ
3sinθ=2-2sin^2θ
2sin^2θ+3sinθ-2=0
(2sinθ-1)(sinθ+2)=0
sinθ=1/2 OR sinθ= -2(rejected)
θ= 30° OR 150°
2.sin²θ-3cosθ-1=0
1-cos^2θ-3cosθ-1=0
cosθ(cosθ+3)=0
cosθ=0 OR cosθ= -3(rejected)
θ= 90° OR 270°
3A)sinx =1/2
x= 150°
B)1-sin²A / cosA =cos^2A/cosA=cosA
參考: ME
收錄日期: 2021-04-23 22:21:54
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