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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Mar 2007 20:19:58 IST
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solve: 5(sin2x - cos2x) = tanx + 5
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4 Mar 2007 13:17:35 IST
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please reply
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There is no better feeling in this world than being a winner! |
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4 Mar 2007 14:03:42 IST
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5(2sinxcosx -(2sin^2x - 1)) = tanx + 5; 10sinx(cosx-sinx) = sinx(1/cosx) ; sinx = 0 or 10(cosx-sinx) = secx ; solve, u'll get the ans.....
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Sorry,i ate ur brain!!! |
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4 Mar 2007 15:47:27 IST
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plz solve!
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4 Mar 2007 16:54:03 IST
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5[2tanx/(1+tan2x) - (1-tan2x)/(1+tan2x)]=tanx +5 5[(2tanx -1 +tan2x-1-tan2x)/(1+tan2x)] = tanx 5(2tanx -2)= tanx +tan3x tan3x -9tanx +10=0 let tanx = t t3 -9t +10=0 let a,b,c be the roots abc = -10 ab+bc+ca = -9 a+b+c=0 solve the 3 equations the roots give the value of tanx from which x can b found and generalised if u found ma soltuion useful plzzzzzzzzz rate me cheeeeeeers
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dbznfreak---watchin episodes for 6 yrs--movin on to dbgt
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4 Mar 2007 18:56:52 IST
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5[2tanx/(1+tan2x) - (1-tan2x)/(1+tan2x)]=tanx +5
how did you get this
?
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4 Mar 2007 22:26:14 IST
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5 Mar 2007 18:50:18 IST
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see,you should be knowing that
sin2x= 2tanx/(1+tan2x) and cos2x=(1-tan2x)/(1+tan2x)
this is easy to prove,as sin2x=2sinxcosx =2(sinx/cosx)cos2x
this equals 2tanx*cos2x = 2tanx/sec2x = 2tanx/(1+tan2 x)!
similarly cos 2x can be found.
on substitution in question and solving,you will get,
tan3 x + 10 - 9tanx=0
the solution will give you the answer!
This is the same as what Vasant has done.
That solves the problem!
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Destiny is what you make
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5 Mar 2007 21:56:53 IST
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now i got it thanks amaron
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8 Mar 2007 18:06:04 IST
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hey miyo 5(2sinxcosx -(2sin^2x - 1)) = tanx + 5; the above step that u typed is wrong cos2x = 1 -2sin^2x and not 2sin^2x-1 proceedin further with ur method we get a very complex equation cheeeers
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dbznfreak---watchin episodes for 6 yrs--movin on to dbgt
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>
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