F5數學一問(三角function)

用戶19771

2012-06-01 5:32 am

1.sin(2x)sin(x)=-1
2.x^2-xsin(x)+1=0

where 0<x<360

請列出步驟
多謝解答

更新1:

對於1) 仲有少少唔明點解2x = 90° and x = 270° or 2x = 270° and x = 90° 咁就冇solution?

回答 (1)

用戶2053

2012-06-01 6:16 am

✔ 最佳答案

1)sin2x sinx = -1
Since -1 ≤ sin2x ≤ 1 and -1 ≤ sinx ≤ 1 , therefore only 2 cases :
sin2x = 1 and sinx = - 1 or sin2x = - 1 and sinx = 1 ,
2x = 90° and x = 270° or 2x = 270° and x = 90° Both cases are no solutions.

Method 2 :
sin2x sinx = -1
2sin²x cosx = - 1
2(1 - cos²x) cosx = - 1
2cos³x - 2cosx - 1 = 0
cos³x - cosx = 1/2
No solutions for - 1 ≤ cosx ≤ 1 ,
So x has no real solutions.

2)x² - xsinx + 1 = 0
x² - (sinx) x + 1 = 0
△= sin²x - 4 ≤ 1² - 4 = - 3 < 0So x has no real roots.

2012-06-01 16:08:05 補充：
2x = 90° and x = 270° (同時)
==>
x = 45° = 270° (荒謬)

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