[tex]\text{tg}(\frac{x}{2}) = \frac{1}{\sqrt{3}}.\\\text{Stiind ca tg$(\frac{\pi}{6})=\frac{1}{\sqrt{3}}$} \implies \frac{x}{2} = \frac{\pi}{6} + 2\pi k, \text{ unde } k \in \mathbb{Z}.\\2x = 4 \cdot \frac{x}{2} = 4 \cdot (\frac{\pi}{6} + 2\pi k) = \frac{2\pi}{3} + 2 \pi (4k) = \frac{2\pi}{3} + 2 \pi k_{1}, \text{ unde } k_{1} = 4k.\\\sin(2x) = 2\sin(x)\cos(x) = \sin(\frac{2\pi}{3} + 2\pi k_{1}) = \frac{\sqrt{3}}{2} => \sin(x)\cos(x) = \frac{\sqrt{3}}{4}.[/tex]
