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![[Post New]](/templates/default/images/icon_minipost_new.gif) 2 Mar 2007 16:07:54 IST
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solve: cos(60-x).cosx.cos(60+x)=1/4
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There is no better feeling in this world than being a winner! |
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2 Mar 2007 17:19:20 IST
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2cos(60-x)cos(60+x) 2cosx=1
[cos(120) + cos(-2x)] 2cosx = 1
[-1/2 + cos 2x] 2cosx = 1
-cosx + 2cosxcos2x = 1
2cosxcos2x - cosx =1
cos 3x + cos x - cos x = 1
cos 3x = 1
Thus 3x = 0
x = 0
x = 2 pi n
I Dont Think I Have Made Any Mistake But If i Have...Do Let Me Know...Rate Me If My Solution Is Adequate
* I Have used cos(-x) = cos x a lot*
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Sometimes One Dream Is Enough To Light Up The Entire October Sky....
First Year Mechanical Engineering
Veermata Jijabai Technological Institute |
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2 Mar 2007 17:19:33 IST
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u want to prove it or find ans
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adversities cause some men to break other to break records............i m of the other type....... :-) |
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2 Mar 2007 17:33:21 IST
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cos(60-x) cosx cos(60+x) = 1/4 [cos260 cos2x - sin260sin2x]cosx =1/4 [(cos2x)/4] cos2x - 3sin2x/4]cosx =1/4 cos4x - 3sin2xcosx =1 cos4x - 3cosx - 3cos3x -1=0 solving the equation u'll get the value of x and then u can generalise it 2npi +-x cheeeeeeeeers cammon rate me.......
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dbznfreak---watchin episodes for 6 yrs--movin on to dbgt
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2 Mar 2007 17:42:21 IST
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the given ans is 2npi/3;n but i think the answer you have given is perfect..take a salute
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2 Mar 2007 17:43:54 IST
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cos4x - 3cosx - 3cos3x -1=0
ok...plz solve it furthur vasanth
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4 Mar 2007 00:58:04 IST
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its a cubic equation in cosx take sum of the roots sum of the product of roots taken 2 at a time product of the roots u'll get three values for cosx and if ur answer of 2npi/3 is correct then the other 2 options will most probably get eliminated i'm in a bit o' hurry i'll complete the solution tom cheeeeeeeeers
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dbznfreak---watchin episodes for 6 yrs--movin on to dbgt
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