數學問題=.=數學問題=.=數學問題啊@@@@@@@@@2

2Cos(3x)Sin(2x) = 0

Case 1: Sin(2x) = 0 -> 2x = nπ -> x = n (π/2)

Case 2: Cos(3x) = 0 -> 3x = (π/2)+nπ -> 3x = (1+2n)(π/2)
-> x = (1+2n)(π/6)

Assume, Cos(3x) = Sin(2x) = 0
4(Cos(x))^3 - 3Cos(x )= 2Sin(x)Cos(x)
4(Cos(x))^2 - 3= 2Sin(x)
4-4(Sin(x))^2 - 3= 2Sin(x)
4(Sin(x))^2 + 2Sin(x) – 1 = 0
Sin(x) = {-2 ± √[4-4*4*(-1)]}/2
Sin(x) = -1 ± √5

Then, (1) Cos(3x) = 0, Sin(2x) ≠ 0
(2) Sin(2x) = 0, Cos(3x) ≠ 0

Thus, either Cos(3x) = 0 or Sin(2x) = 0

2009-02-13 02:13:51 補充：
(1) Cos(3x) = 0, Sin(2x) not equal to 0

(2) Sin(2x) = 0, Cos(3x) not equal to 0

因為 2cos3x = 1 嘛

2009-02-13 18:09:36 補充：
不用想得那麼複雜的

2cos3xsin2x = 0 sin2x = 0 好像應該討論 cos3x=0情況
修改題目順序 sin2x = 0 2cos3xsin2x = 0
很明顯在已知sin2x=0情況下2cos3xsin2x = 0是必然事實
因此題目剩下一條件sin2x=0
sin2x=0 則 2x=n*(pei) n 為整數 x=n*(pei) n 為整數

2009-02-13 09:43:08 補充：
最後一行修改如下
sin2x=0 則 2x=n*(pei) n 為整數 x=n*(pei)/2 n 為整數