If 0∘≦ x ≦360∘,what is the range for sin2x?
If 0∘≦ x ≦360∘,what is the number of roots of the equation 2sinx cosx =sinx?
(Please explain the answer as well, thank you)
F.4 Maths
2007-06-12 4:29 am
回答 (2)
2007-06-12 4:48 am
✔ 最佳答案
For the first one, -1<=siny<=1Let y=2x
-1<=sin2x<=1
sin any number 都係between -1 and 1
For the second one, 2sinxcosx= sinx
2sinxcosx-sinx=0
sinx(2cosx-1)=0
sinx =0 or 2cosx-1 =0
x= 0∘,180∘,360∘ or cosx =1/2
x= 60∘, 300∘
so x= 0∘,60∘,180∘, 300∘or 360∘
2007-06-11 20:49:39 補充：
上面果位人兄差個180度
2007-06-11 20:50:34 補充：
第一句 係For the first one, -1=< 'siny '=< 1
2007-06-12 4:45 am
1)
-1≦sin2x≦1.
∵For any complete cycle of x (0°to360°).
2)
2sinx cosx –sinx =0
sinx (2cosx –1)=0
sinx=0 or cosx =1/2
x = 0° or 360° or 60° or 300°.
∴4 roots.
-1≦sin2x≦1.
∵For any complete cycle of x (0°to360°).
2)
2sinx cosx –sinx =0
sinx (2cosx –1)=0
sinx=0 or cosx =1/2
x = 0° or 360° or 60° or 300°.
∴4 roots.
參考: ki
收錄日期: 2021-04-13 14:31:31
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