Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 62

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

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

\[2x = ( - 1)^{n + 1} \bullet \frac{\pi}{6} + \pi n\]

\[x = ( - 1)^{n + 1} \bullet \frac{\pi}{12} + \frac{\pi n}{2}.\]

\[Ответ:\ - \frac{17\pi}{12};\ - \frac{13\pi}{12};\ - \frac{5\pi}{12};\ \]

\[- \frac{\pi}{12};\ \frac{7\pi}{12};\ \frac{11\pi}{12}.\]

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

\[3x = ( - 1)^{n} \bullet \frac{\pi}{3} + \pi n\]

\[x = ( - 1)^{n} \bullet \frac{\pi}{9} + \frac{\pi n}{3}.\]

\[Ответ:\ - \frac{11\pi}{9};\ - \frac{10\pi}{9};\ - \frac{5\pi}{9};\ \]

\[- \frac{4\pi}{9};\ \frac{\pi}{9};\ \frac{2\pi}{9};\ \frac{7\pi}{9};\ \frac{8\pi}{9}.\]