Răspuns:
[tex] x \in \Big\{\frac{5\pi}{12}, \frac{3\pi}{4}, \frac{7\pi}{4}, \frac{17\pi}{12}\Big\}[/tex]
Explicație pas cu pas:
[tex]\sin{\alpha} = -\frac{1}{2} \implies \alpha \in \{210^\circ, 330^\circ\}\\ 210^\circ + 360^\circ\cdot k= 210\cdot \frac{\pi}{180} = \frac{7\pi}{6} + 2k\pi, k\in\mathbb{Z}\\ \\ 330^\circ+360^\circ\cdot k= 330\cdot \frac{\pi}{180} = \frac{11\pi}{6} + 2k\pi, k\in \mathbb{Z}\\ \\ \alpha = 2x+\frac{\pi}{3} = 2x+\frac{2\pi}{6}\\ \\ \alpha = \frac{7\pi}{6}+2k\pi \\ \\ 2x+\frac{2\pi}{6} = \frac{7\pi}{6}+2k\pi \\ \\ 2x = \frac{5\pi}{6}+2k\pi \\ \\ x = \frac{5\pi}{12} + k\pi \implies x_1 = \frac{5\pi}{12}, x_2 = \frac{17\pi}{12}, x_3 = \frac{29\pi}{12}\cdots \\ \\ \textrm{Deoarece } x \in (0, 2\pi)\\ \\ x \in \Big\{\frac{5\pi}{12}, \frac{17\pi}{12}\Big\}\\ \\ \alpha = \frac{11\pi}{6} + 2k\pi \\ \\ 2x+\frac{2\pi}{6} = \frac{11\pi}{6} + 2k\pi \\ \\ 2x = \frac{9\pi}{6} + 2k\pi = \frac{3\pi}{2} + 2k\pi\\ \\ x = \frac{3\pi}{4} + k\pi\\ \\ x_1 = \frac{3\pi}{4}, x_2 = \frac{7\pi}{4} [/tex]
