Odpowiedź:
[tex]\sin\alpha + \sin\beta = 2\sin\frac{\alpha + \beta}{2}\cos\frac{\alpha-\beta}{2}[/tex]
[tex]\sin\left(x+\frac{\pi}{6}\right) + \sin\left(x-\frac{\pi}{6}\right) = 2\sin\left(\frac{x+\frac{\pi}{6}+x-\frac{\pi}{6}}{2}\right)\cos\left(\frac{x+\frac{\pi}{6}-(x-\frac{\pi}{6})}{2}\right) = 2\sin(x)\cos\left(\frac{\pi}{6}\right) = 2\sin(x) \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \sin x[/tex]
[tex]\sqrt{3} \sin x = \frac{1}{2}[/tex]
[tex]\sin x = \frac{\sqrt{3}}{6}[/tex]
Dla [tex]x \in (0, 2\pi)[/tex]:
[tex]x = \arcsin\left( \frac{\sqrt{3}}{6} \right) \vee x = \pi - \arcsin\left( \frac{\sqrt{3}}{6} \right)[/tex]