Ukrajina 2008

Fig.8

The graph of functions $y=|x-a|+|x-b|$ for $b>a$ is shown in fig.8. Thus it is clear that equation $|x-a|+|x-b|=c$ may have a single solution only if $a=b$ and $c=0$ (fig.9).

Therefore for our equation the following condition should be satisfied: $$\{\begin{array}{l}\sin \alpha =\frac{1}{2}\\ \cos 3\alpha =0\end{array}\Rightarrow \{\begin{array}{ll}\alpha =(-1{)}^n\frac{\pi }{6}+\pi n,& n,k\in \mathbb{Z}\\ \alpha =\frac{\pi }{6}+\frac{1}{3}\pi k& \end{array}$$ This system has the following common solution: $\alpha =(-1{)}^n\frac{\pi }{6}+\pi n,n\in \mathbb{Z}$ and $x=\frac{1}{2}$.

Fig.9