IIT JEE , AIEEE Forum - Mathematics , Trignometry

rahulraj1

If e^sinX-e^-sinX=4 then find the number of real solutions?

nishantsingh89

number of real solutions is 0

y= e^sinx

then you can write this expression as

y^2-4y-1=0

y= 2+sqrt(5) or y=2-sqrt(5)

now 1/e <e^sinx<e { -1<sinx<1}

and both the above obtained values lie outside the specified range

dream

put e^sinx =t

then by rearranging the given equation we get,
t^2-4t-1=0

so t= 2+root5 or 2-root5
= 4.43 or -.23

but t= e^sinx
hence sin x = ln t
also log of -ve no is not defined
and ln 4.43= 1.44

as log of -ve no is not defined and sinx lies between -1 and 1,

the given equation has no real roots

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