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2008-02-15 10:25 pm
1.sin2X-√3 cosx'=0

2.cos6x/cos2x+sin6x/sin2x =4 cos2x

Please give me the solution
THANKS SO MUCH!!!!

回答 (2)

2008-02-15 10:42 pm
✔ 最佳答案
Solution:

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2008-02-15 10:51 pm
sin2X-√3 cosx=0
2sinx cosx -√3 cosx=0
cosx(2sinx-√3)=0
cosx=0 or sinx=√3/2
cosx=cos 90 or 270 or sinx = sin 60 or sin 120
so x= 60 , 90 , 120 ,270//

cos6x/cos2x+sin6x/ sin2x =4 cos2x
(cos6x sin2x + sin6x cos2x)/(sin2x cos 2x)=4cos2x
sin8x=4cos2x (sinx cosx)
sin8x=2sin2x cos2x
sin8x=sin4x
2cos4x sin4x-sin4x=0
(sin4x)(2cos 4x -1)=0
sin4x=0 or cos4x = 1/2
sin4x= sin 0 ,sin 180 , sin360 or cos4x= cos cos 60 or 300
4x= 0 , 60 , 180 , 300 or 360
x=0 ,15 , 45 , 75 or 90//


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