求下列各方程由0至2π的解,答案須保留π
4sin²x+3cot²x=4
a-maths!!!
2007-12-05 2:00 am
回答 (1)
2007-12-05 2:40 am
✔ 最佳答案
Note:Gabriella Montez~永不言敗!!!在最後的答案不應該有π/1的。
4sin²x+3cot²x = 4
4sin4x+3cos²x = 4sin²x
4sin4x+3(1-sin²x) = 4sin²x
4sin4x – 7 sin²x +3 = 0
sin²x = 1 or 3/4
sin x = ±1 or ±(sqrt3)/2
x = π/2 or 3π/2 or π/3 or 2π/3 or 4π/3 or 5π/3.
收錄日期: 2021-04-23 20:35:02
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071204000051KK01978