3. In $\triangle A B C$, $\angle B=\frac{\pi}{3}$, point $D$ is on side $A B$, $B D=1, A C=\sqrt{3}, D A=D C$. Then $\angle D C A=$ $\qquad$
3. $\frac{\pi}{6}$ or $\frac{\pi}{18}$.

In $\triangle A C D$, since $D A=D C$, we can set $\angle A=\angle D C A=\theta$. Then $\angle A D C=\pi-2 \theta$.

Given $A C=\sqrt{3}$, by the Law of Sines, we have
$$
\frac{A C}{\sin 2 \theta}=\frac{C D}{\sin \theta} \Rightarrow C D=\frac{\sqrt{3}}{2 \cos } \bar{\theta} \text {. }
$$

In $\triangle B D C$,
$$
\angle B D C=2 \theta, \angle B C D=\frac{2 \pi}{3}-2 \theta \text {. }
$$

Combining the Law of Sines, we get
$$
\begin{array}{l}
\frac{C D}{\sin B}=\frac{B D}{\sin \angle B C D} \\
\Rightarrow \frac{\frac{\sqrt{3}}{2 \cos \theta}}{\sin \frac{\pi}{3}}=\frac{1}{\sin \left(\frac{2 \pi}{3}-2 \theta\right)} \\
\Rightarrow \sin \left(\frac{\pi}{2}-\theta\right)=\sin \left(\frac{2 \pi}{3}-2 \theta\right) .
\end{array}
$$

Since $0<\theta<\frac{\pi}{2}$, we have
$$
\begin{array}{l}
0<\frac{\pi}{2}-\theta<\frac{\pi}{2},-\frac{\pi}{3}<\frac{2 \pi}{3}-2 \theta<\frac{2 \pi}{3} . \\
\text { Therefore, } \frac{\pi}{2}-\theta=\frac{2 \pi}{3}-2 \theta \text { or } \frac{\pi}{2}-\theta+\frac{2 \pi}{3}-2 \theta=\pi \\
\Rightarrow \theta=\frac{\pi}{6} \text { or } \frac{\pi}{18} \\
\Rightarrow \angle D C A=\frac{\pi}{6} \text { or } \frac{\pi}{18} .
\end{array}
$$