The equation $e^{\sin x}-e^{-\sin x}-4=0$ has:

Given equation is  $e^{\sin x}-e^{-\sin x}-4=0$ 

 put $e^{\sin x}=t$ in the given equation,

 we get

 $t^2-4t-1=0$ 

$\Rightarrow t=\frac{4\pm 2\sqrt{5}}{2}=2\pm \sqrt{5}$            

$\Rightarrow e^{\sin x}=2\pm \sqrt{5} (\because t=e^{\sin x})$

$\Rightarrow e^{\sin x}=2-\sqrt{5}$ and $e^{\sin x}=2+\sqrt{5}$

$e^{\sin x}=2-\sqrt{5}<0 is not possible$

$\therefore e^{\sin x}=2+\sqrt{5}$

 $$\begin{gathered} \Rightarrow \sin x=\ln \left(2+\sqrt{5}\right)>1 \\ This is not true. \\ \therefore There is no solution for the given equation. \end{gathered}$$