ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 64

\[- \frac{3\pi}{2} \leq x \leq \pi.\]

\[1)\sin{2x} \geq - \frac{1}{2}\]

\[- \frac{\pi}{6} + 2\pi n \leq 2x \leq \frac{7\pi}{6} + 2\pi n\]

\[- \frac{\pi}{12} + \pi n \leq x \leq \frac{7\pi}{12} + \pi\text{n.}\]

\[Ответ:\ - \frac{3\pi}{2} \leq x \leq - \frac{17\pi}{12};\ \ \ \]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ - \frac{13\pi}{12} \leq x \leq - \frac{5\pi}{12};\ \ \ \]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - \frac{\pi}{12} \leq x \leq \frac{7\pi}{12};\ \ \ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\frac{11\pi}{12} \leq x \leq \pi.\]

\[2)\sin{3x} < \frac{\sqrt{3}}{2}\]

\[\frac{2\pi}{3} + 2\pi n < 3x < \frac{7\pi}{3} + 2\pi n\]

\[\frac{2\pi}{9} + \frac{2\pi n}{3} < x < \frac{7\pi}{9} + \frac{2\text{πn}}{3}.\]

\[Ответ:\ - \frac{3\pi}{2} \leq x < - \frac{11\pi}{9};\ \ \]

\[- \frac{10\pi}{9} < x < - \frac{5\pi}{9};\ \]

\[- \frac{4\pi}{9} < x < \frac{\pi}{9};\ \]

\[\ \frac{2\pi}{9} < x < \frac{7\pi}{9};\ \ \ \frac{8\pi}{9} < x \leq \pi.\]