Show that the equation `e^(sinx)-e^(-sinx)-4=0` has no real solution.

Options

A. exactly one real root
B. exactly one real root
C. exactly four real roots
D. infinite number of real roots

Answer

D

Explanation

Lert `e^(sin x)=t`
Then lie given equation can be written as
`t-1/t-4=0impliest^(2)-4t-1=0`
`:.t=(4+sqrt((16+4)))/2`
`impliese^(sinx)=(2+sqrt(5))[ :' e^(sinx)gt0,:` taking `+ve` sign]
`impliessinx=log_(e)(2+sqrt(5))`………….ii
`:'(2+sqrt(5))gte[:' e=2.71828]`
`implieslog_(e)(2+sqrt(5))gt1` ............iii
From eq. ii and iii we get
`sin x gt1` [ which is impossible]`
Hence no real root exists.