3. Given
$$
\sin \alpha+\sqrt{3} \sin \beta=1, \cos \alpha+\sqrt{3} \cos \beta=\sqrt{3} \text {. }
$$

then the value of $\cos (\alpha-\beta)$ is $\qquad$
3. 0 .

Squaring and adding the two known equations, we get
$$
\begin{array}{l}
4+2 \sqrt{3}(\sin \alpha \cdot \sin \beta+\cos \alpha \cdot \cos \beta)=4 \\
\Rightarrow \cos (\alpha-\beta)=0
\end{array}
$$