ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 59

\[\lbrack 0;\ 3\pi\rbrack.\]

\[1)\sin x > \frac{1}{2}\]

\[\frac{\pi}{6} + 2\pi n < x < \frac{5\pi}{6} + 2\pi\text{n.}\]

\[Ответ:\ \ \frac{\pi}{6} < x < \frac{5\pi}{6};\ \ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ }\frac{13\pi}{6} < x < \frac{17\pi}{6}.\]

\[2)\sin x \leq \frac{\sqrt{2}}{2}\]

\[\frac{3\pi}{4} + 2\pi n \leq x \leq \frac{9\pi}{4} + 2\pi\text{n.}\]

\[Ответ:\ \ 0 \leq x \leq \frac{\pi}{4};\ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\frac{3\pi}{4} \leq x \leq \frac{9\pi}{4};\ \ \ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ }\frac{11\pi}{4} \leq x \leq 3\pi.\]

\[3)\sin x \geq - \frac{1}{2}\]

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

\[Ответ:\ \ 0 \leq x \leq \frac{7\pi}{6};\ \ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ }\frac{11\pi}{6} \leq x \leq 3\pi.\]

\[4)\sin x < - \frac{\sqrt{3}}{2}\]

\[\frac{4\pi}{3} + 2\pi n < x < \frac{5\pi}{3}.\]

\[Ответ:\ \ \frac{4\pi}{3} < x < \frac{5\pi}{3}.\]