3. The solution to the equation $\sqrt{13-\sqrt{13+x}}$ $=x$ is $\qquad$
3.3.

Let $\sqrt{13+x}=y$ then
$$
\left\{\begin{array}{l}
\sqrt{13-y}=x, \\
\sqrt{13+x}=y .
\end{array}\right.
$$
(2) $)^{2}-(1)^{2}$ gives $(x+y)(x-y+1)=0$.

From the original equation, we know $x>0$, and from (2), $y>0$, so $x+y \neq 0$. Thus, $x-y+1=0$, which means $y=x+1$. Therefore, it is easy to get $x=3$ (discard $x=-4$).