2.  The solution set of the inequality $\sqrt{x^{2}+4 x} \leqslant 4-\sqrt{16-x^{2}}$ is $\qquad$ .
2. $\{-4,0\}$.

For the expression under the square root to be meaningful in the inequality, we have
$$
\left\{\begin{array}{l}
x^{2}-4 x \geqslant 0, \\
16-x^{2} \geqslant 0 .
\end{array}\right.
$$

This gives $x=-4$ or $x \in[0,4]$.
Clearly, $x=-4$ is a solution to the inequality. For $x \in[0,4]$, we have
$$
\begin{aligned}
4 & \geqslant \sqrt{x^{2}+4 x}+\sqrt{16-x^{2}} \\
& =\sqrt{x+4}(\sqrt{x}+\sqrt{4-x}) \\
& \geqslant 2(\sqrt{x}+\sqrt{4-x}) \\
& =2 \sqrt{(\sqrt{x}+\sqrt{4-x})^{2}} \\
& =2 \sqrt{4+2 \sqrt{x(4-x)}} \\
& \geqslant 2 \cdot \sqrt{4}=4
\end{aligned}
$$
Thus, for $x \in[0,4]$,
$$
\begin{aligned}
& 1 / \sqrt{x^{2}+4 x}+\sqrt{16-x^{2}} \\
= & \sqrt{16+4 x+2 \sqrt{\left(x^{2}+4 x\right)\left(16-x^{2}\right)}} \\
\geqslant & \sqrt{16}=4
\end{aligned}
$$

This leads to $x=0$.