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Algebra

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Joined:	1 May 2007
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4 Jul 2007 22:46:07 IST

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number of roots

None

Number of real roots of the equation e^sinx- e^-sinx-4 = 0 is ______

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karan bhatia

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Joined: 21 Jun 2007

Posts: 130

4 Jul 2007 23:07:42 IST

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the number of roots r 0
coz if we replace e^sinx by y
then the roots cumin r not in the range of e^sinx

rate me if it was helpful

geethu

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Joined: 7 May 2007

Posts: 193

5 Jul 2007 08:37:46 IST

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e^sinx - e^-sinx-4 =0

e^2sinx -4e^sinx - 1 = 0

e^sinx = 2

but e^x>0 4 any x ie e^sinx = 2+

sin x=log(2+

5) >1 {e lies in (2,3) }

ie no of real sol=0 {sinx lies in [-1,1] }

soham chokshi

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Joined: 2 Jul 2007

Posts: 30

5 Jul 2007 08:54:38 IST

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haha what a coinci !!!
i solved the same sum just 2 days bak !!!

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