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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: Solve de equation.......
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[Post New]5 Jul 2008 14:12:39 IST
    Subject: Solve de equation.......
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yash_gryffindor (609)

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ecosx = e-cosx+4
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[Post New]5 Jul 2008 14:15:21 IST
    Subject: Re:Solve de equation.......
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sachinguptaiit (940)

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(ecosx)2=1+4ecosx



 


put ecosx=t



 


t2-4t-1=0



 


t=2\pm \sqrt{5}



 


Since t>0 we have ecosx= 2+\sqrt{5}



 


hence cosx=log(2+\sqrt{5})



 


 

but cosx belongs to -1 and 1


and ln(2+\sqrt{5})>1 hence no solution


edit:-sorry for mistake
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[Post New]5 Jul 2008 14:20:13 IST
    Subject: Re:Solve de equation.......
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allamraju (3422)

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Put ecosx=t,then t=1/t+4

So,t2-4t-1=0(t-2)2=5

t=2+rt5 or 2-rt5

But ecosx is >0 for all x,so,t=ecosx=2+rt5

So,cosx=ln(2+rt5)>ln4>1 and hence,No x exists.

So,the eqn. has no solution.
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