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

Let $t=e^{\sin x}$. Since $-1\le \sin x\le 1$, the range of $t$ is $[\frac{1}{e},e]$.The given equation becomes:
$$t-\frac{1}{t}-4=0$$
$$t^2-4t-1=0$$
Using the quadratic formula:
$$t=\frac{4\pm \sqrt{16+4}}{2}=\frac{4\pm \sqrt{20}}{2}=2\pm \sqrt{5}$$
The two possible values for $t$ are:$t_1=2+\sqrt{5}\approx 2+2.236=4.236$$t_2=2-\sqrt{5}\approx 2-2.236=-0.236$ Checking the range of $t\in [\frac{1}{e},e]\approx [0.368,2.718]$:$t_1=4.236>e$, so it is not possible.$t_2=-0.236<\frac{1}{e}$, so it is not possible.Thus, there are no real values of $x$ that satisfy the equation.