Given that the equation
3x^2-(4sinθ)x+2(1-cosθ)=0
in x has two distinct real roots, and θis an acute angle.Find the range of values of θ
f.4 amaths
2007-03-15 6:43 am
回答 (3)
2007-03-15 7:08 am
✔ 最佳答案
Please find the deductions as follows:圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Crazytrigo67.jpg
參考: My Maths knowledge
2007-03-15 7:11 am
delta > 0
(-4sinθ)^2 - 4(3)(2(1-cosθ)) > 0
16sin^2 θ - 24 + 24cosθ > 0
16(1 - cos^2 θ) - 24 + 24cosθ > 0
16cos^2 θ - 24 cosθ + 8 < 0
2cos^2 θ - 3 cosθ + 1 < 0
(2 cosθ - 1)(cosθ -1 ) < 0
1/2 < cosθ < 1 and 0 <= θ <= 90
cos 60 < cos θ < cos 0 and 0 <= θ <= 90
So, 0 < θ < 60
(-4sinθ)^2 - 4(3)(2(1-cosθ)) > 0
16sin^2 θ - 24 + 24cosθ > 0
16(1 - cos^2 θ) - 24 + 24cosθ > 0
16cos^2 θ - 24 cosθ + 8 < 0
2cos^2 θ - 3 cosθ + 1 < 0
(2 cosθ - 1)(cosθ -1 ) < 0
1/2 < cosθ < 1 and 0 <= θ <= 90
cos 60 < cos θ < cos 0 and 0 <= θ <= 90
So, 0 < θ < 60
2007-03-15 7:04 am
If the equation has 2 distinct real roots,
Discriminant > 0
[-(4sinθ)]^2 - 4 (3) [2(1-cosθ)] > 0
16sin^2θ - 24 + 24cosθ > 0
16(1 - cos^2θ) - 24 + 24cosθ > 0
16 - 16cos^2θ - 24 + 24cosθ > 0
2 - 2cos^2θ - 3 + 3cosθ > 0
2cos^2θ - 3cosθ + 1 > 0
(2cosθ - 1)(cosθ - 1) > 0
1/2 < cosθ < 1
0 < θ < 60
Discriminant > 0
[-(4sinθ)]^2 - 4 (3) [2(1-cosθ)] > 0
16sin^2θ - 24 + 24cosθ > 0
16(1 - cos^2θ) - 24 + 24cosθ > 0
16 - 16cos^2θ - 24 + 24cosθ > 0
2 - 2cos^2θ - 3 + 3cosθ > 0
2cos^2θ - 3cosθ + 1 > 0
(2cosθ - 1)(cosθ - 1) > 0
1/2 < cosθ < 1
0 < θ < 60
收錄日期: 2021-04-12 18:46:11
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070314000051KK04957