How to solve the following equations for 0 <= θ <= 2θ
6 tan^2 θ - 4 sin ^2 θ = 1
????
f.4 a.maths
2007-02-17 10:48 pm
回答 (2)
2007-02-17 11:51 pm
✔ 最佳答案
Please find the deductions below:圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Gensol10.jpg
參考: My Maths knowledge
2007-02-17 11:47 pm
6 tan² θ - 4 sin² θ = 1
6 (sec² θ - 1) - 4 (1 - cos² θ) = 1
6 sec² θ - 11 + 4 cos² θ = 0
(2 sec θ - cos θ) (3 sec θ - 4 cos θ) = 0
2 sec θ = cos θ or 3 sec θ = 4 cos θ
cos² θ = 2 or cos² θ = 3/4
no solution or cos θ = ±(√3)/2
For 0≤θ≤2π,
θ = π/6, π - π/6, π + π/6, 2π - π/6
= π/6, 5π/6, 7π/6, 11π/6.
2007-02-19 00:34:05 補充：
我的答案比第二位的答案要早, 而且結果一樣, 為什麼會選了第二位為最佳答案?
6 (sec² θ - 1) - 4 (1 - cos² θ) = 1
6 sec² θ - 11 + 4 cos² θ = 0
(2 sec θ - cos θ) (3 sec θ - 4 cos θ) = 0
2 sec θ = cos θ or 3 sec θ = 4 cos θ
cos² θ = 2 or cos² θ = 3/4
no solution or cos θ = ±(√3)/2
For 0≤θ≤2π,
θ = π/6, π - π/6, π + π/6, 2π - π/6
= π/6, 5π/6, 7π/6, 11π/6.
2007-02-19 00:34:05 補充：
我的答案比第二位的答案要早, 而且結果一樣, 為什麼會選了第二位為最佳答案?
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https://hk.answers.yahoo.com/question/index?qid=20070217000051KK01874