A.Maths...General Solution...10分....趕急~~

2008-12-13 5:47 am
Answer ALL Questions.

1)Find the general solution of the equation 2tanθ = 3cosecθin radians.

2)Find the gereral solution of the equation 2cos²x - 5sinxcosx + 5sin²x = 1 in radians. Give your answer correct to 2 decimal places.

3)Find the general solution of the following equations in radians.
(a) sinθ - sin2θ + sin3θ=0
(b)sin3θsinθ = sin7θsin5θ
(c)sinθsin7θ + cos3θcos5θ =0
更新1:

Numeral Answer:: 1)2nπ± π/3 2)n +0.24 , nπ +0.79 3)(a) (nπ)/2 , 2nπ± π/3 (b)(nπ)/8 (c)(nπ)/2±π/2 , nπ±π/4

回答 (2)

2008-12-13 6:12 am
✔ 最佳答案
As follow AS~~~~

圖片參考:http://e.imagehost.org/0150/ScreenHunter_02_Dec_12_10_20.gif


2008-12-12 22:23:04 補充:
2)
http://e.imagehost.org/0795/ScreenHunter_03_Dec_12_10_30.gif

2008-12-12 22:46:21 補充:
3)
http://e.imagehost.org/0102/ScreenHunter_04_Dec_12_10_55.gif

2008-12-13 22:59:12 補充:
Corr

3(b) (c)
http://e.imagehost.org/0844/ScreenHunter_06_Dec_13_11_07.gif
2008-12-13 8:17 am
There is a mistake in #001 solution, Q3b:

sin8θ=0 => 8θ=nπ but not 2nπ
sin4θ=0 => 4θ=nπ but not 2nπ

Therefore θ=nπ/8 (which already include θ=nπ/4)

2008-12-13 00:20:56 補充:
Also in Q3c, there is also a mistake:

cos4θ=0 => 4θ=2nπ±π/2 but not 2nπ
cos2θ=0 => 2θ=2nπ±π/2 but not 2nπ

Therefore θ=nπ/2±π/8 or θ=nπ±π/4


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