Start with the given equation: $e^{\sin x} - e^{-\sin x} - 4 = 0$ Rearranging, $e^{\sin x} - e^{-\sin x} = 4$

Let $t = \sin x$ Since $x$ is real, we know $-1 \le t \le 1$

So the equation becomes $e^t - e^{-t} = 4$

Define $f(t) = e^t - e^{-t} = 2\sinh t$ We now check the possible values of $f(t)$ for $t \in [-1,1]$.

Since $f(t)$ is increasing, its maximum on $[-1,1]$ occurs at $t=1$: $f(1) = e - e^{-1}$ Numerically, $e - e^{-1} \approx 2.718 - 0.368 = 2.350$

Its minimum occurs at $t=-1$: $f(-1) = e^{-1} - e = -2.350$

Hence, $f(t) \in [-(e-e^{-1}),\, e-e^{-1}] \approx [-2.350, 2.350]$

But the equation requires $f(t)=4$ which is impossible because $4$ is outside the range of $f(t)$.

Therefore, there is no real value of $t=\sin x$ satisfying the equation, and hence no real $x$ satisfies it.

So the equation has: $\boxed{\text{no real roots}}$

Comparing with the stored correct answer: Stored answer = B

Our derived answer is also B, so they agree.