The equation $\mathrm{e}^{\sin x}-\mathrm{e}^{-\sin x}=4$ has _______ solutions.

$\mathrm{e}^{\sin x}-e^{-\sin x}-4=0$
Let $\mathrm{e}^{\sin x}=y$, then the given equation can be written as $y^2-4 y-1=0 \Rightarrow y=2 \cdot \pm \sqrt{5}$ But the value of $y=\mathrm{e}^{\sin x}$ is always positive, so $\begin{aligned} & y=2+\sqrt{5} \\ & \Rightarrow \mathrm{e}^{\sin x}=2+\sqrt{5} \\ & \Rightarrow \mathrm{e}^{\sin x}\gt\mathrm{e} \\ & \Rightarrow \sin x\gt1 \end{aligned}$ $\begin{aligned} & \ldots[\because 2 \lt \sqrt{5}] \\ & \ldots[\because 2+\sqrt{5}\gt\mathrm{e}] \end{aligned}$ which is not possible, since $\sin x$ cannot be greater than 1. Hence, no real value of $x$ satisfies the given equation.