(4)2sin4xcosx+sin7x=0
(5)1+secx=【cot(x/2)】^2
(6)sinx+sin2x+sin3x=1+cosx+cos2x
答案
(4)(kπ/5,kπ±π/3)
(5)(2kπ+π,2kπ±π/3)
(6)(2kπ/3+π/6,2kπ+π/2,2kπ±2π/3);
(kπ+π/2,2kπ±2π/3,kπ+π/6(-1)^k)
答案不一定都正確哦!!!
有可能錯!!!
不知道這種題目對大家來說有沒有難度...對我而言...可是很難的!!!
三角方程(2)...難!!!有答案!!!要過程...
2007-07-15 3:16 am
回答 (2)
2007-07-15 5:01 am
✔ 最佳答案
As follows:圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Jul07/Crazytrigo11.jpg
圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Jul07/Crazytrigo12.jpg
圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Jul07/Crazytrigo13.jpg
參考: My Maths knowledge
2007-07-15 5:08 am
(4)2sin4xcosx+sin7x=0
2[1/2 (sin5x+sin3x)]+sin7x=0
sin3x+sin5x+sin7x=0
2sin5xcos-2x +sin5x=0
sin5x(2cos2x+1) =0
sin5x=0 或 cos2x=-1/2
5x=kπ+(-1)^k 0 或 2x=2kπ± 2π/3
x=kπ/5 或 x=kπ± π/3
sor.....其他冇時間諗....下次再答你
2[1/2 (sin5x+sin3x)]+sin7x=0
sin3x+sin5x+sin7x=0
2sin5xcos-2x +sin5x=0
sin5x(2cos2x+1) =0
sin5x=0 或 cos2x=-1/2
5x=kπ+(-1)^k 0 或 2x=2kπ± 2π/3
x=kπ/5 或 x=kπ± π/3
sor.....其他冇時間諗....下次再答你
收錄日期: 2021-04-29 19:40:53
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