6. Let $\alpha \in(0, \pi)$, the equation $\left|2 x-\frac{1}{2}\right|+|(\sqrt{6}-\sqrt{2}) x|=\sin \alpha$ has exactly one real root, then $\alpha=$
Solve: $\frac{\pi}{12}$ or $\frac{11 \pi}{12}$.
$$
\text { Let } f_{1}(x)=\left|2 x-\frac{1}{2}\right|+|(\sqrt{6}-\sqrt{2}) x|, f_{2}(x)=\sin \alpha
$$

As shown in the figure, it is easy to see that the graph of $f_{1}(x)$ is a broken line, and $f_{2}(x)$ is a straight line parallel to the $x$-axis. If the equation $\left|2 x-\frac{1}{2}\right|+|(\sqrt{6}-\sqrt{2}) x|=\sin \alpha$ has exactly one real root, then
$$
f_{2}(x)=f_{1}(x)_{\min }=\min \left\{f_{1}(0), f_{1}\left(\frac{1}{4}\right)\right\}=\min \left\{\frac{1}{2}, \frac{\sqrt{6}-\sqrt{2}}{4}\right\}=\frac{\sqrt{6}-\sqrt{2}}{4}
$$

Therefore, $\sin \alpha=\frac{\sqrt{6}-\sqrt{2}}{4}$, and $\alpha \in(0, \pi)$, so $\alpha=\frac{\pi}{12}$ or $\frac{11 \pi}{12}$.