نکته: $\operatorname{sina}+\operatorname{sinb}=2\sin \frac{a+b}{2}\cos \frac{a-b}{2}$  با توجه به نکتۀ بالا، می نویسیم: $\left( \operatorname{sina}+\sin \left( \frac{\pi }{3}+a \right) \right)+\sin \left( \frac{\pi }{6}+a \right)=\frac{3+\sqrt{3}}{2}$  $\Rightarrow 2\sin \left( \frac{\pi }{6}+a \right)\cos \frac{\pi }{6}+\sin \left( \frac{\pi }{6}+a \right)$  $=\sqrt{3}\sin \left( \frac{\pi }{6}+a \right)+\sin \left( \frac{\pi }{6}+a \right)=\left( \sqrt{3}+1 \right)\sin \left( \frac{\pi }{6}+a \right)=\frac{3+\sqrt{3}}{2}$  $\Rightarrow \left( \sqrt{3}+1 \right)\sin \left( \frac{\pi }{6}+a \right)=\frac{\sqrt{3}\left( \sqrt{3}+1 \right)}{2}\xrightarrow{\div \left( \sqrt{3}+1 \right)}\sin \left( \frac{\pi }{6}+a \right)=\frac{\sqrt{3}}{2}=\sin \frac{\pi }{3}$  $\Rightarrow \left\{ \begin{matrix}    \frac{\pi }{6}+a=2k\pi +\frac{\pi }{3}\Rightarrow a=2k\pi +\frac{\pi }{6}\xrightarrow{k=0}a=\frac{\pi }{6}  \\    \frac{\pi }{6}+a=2k\pi +\frac{2\pi }{3}\Rightarrow a=2k\pi +\frac{\pi }{2}\xrightarrow{k=0}a=\frac{\pi }{2}  \\ \end{matrix} \right.$  ئس معادله در بازۀ $\left[ 0,2\pi  \right]$ تنها 2 جواب دارد.