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

Let $\angle A=\angle D C A=\theta \in\left(0, \frac{\pi}{2}\right)$.
In $\triangle A D C$, using the Law of Sines, we get
$$
\frac{A C}{\sin 2 \theta}=\frac{C D}{\sin \theta} \Rightarrow C D=\frac{\sqrt{3}}{2 \cos \theta} \text {. }
$$

Similarly, in $\triangle B C D$, using the Law of Sines and combining with $C D$, we get
$$
\cos \theta=\sin \left(\frac{2 \pi}{3}-2 \theta\right) \Rightarrow \theta=\frac{\pi}{6}, \frac{\pi}{18} \text {. }
$$