$$
\text { Given that, } e^{\sin x}-e^{-\sin x}-4=0
$$
$$
\begin{aligned}
&\begin{array}{l}
\text { Let } e^{\sin x}=t \\
\therefore \quad t-t^{-1}-4=0 \\
\Rightarrow \quad t-\frac{1}{t}-4=0 \\
\Rightarrow \quad t^{2}-1-4 t=0 \\
\quad t=\frac{4 \pm 2 \sqrt{5}}{2}=2 \pm \sqrt{5} \\
\Rightarrow \quad e^{\sin x}=2 \pm \sqrt{5} \\
\Rightarrow \quad \quad \quad e^{\sin x}=2+\sqrt{5} \text { and } e^{\sin x}=2-\sqrt{5} \\
\Rightarrow \quad e^{\sin x}=2 + \sqrt{5} \\
\text { Also, }-1 \leq \sin x \leq 1 \\
\therefore \frac{1}{e} \leq e^{\sin x} \leq e \\
\therefore e^{\sin x} \neq 2+\sqrt{5}
\end{array}\\
&\text { Hence, there is no value of } x \text { . }
\end{aligned}
$$