Задание
Реши уравнение \(\sin\left(-\frac{3\pi}{2} + 6x\right) = \frac{\sqrt{2}}{2}\) и выбери корни из отрезка \(\left[-\frac{3\pi}{2}; -\frac{5\pi}{6}\right]\).
(Корни расположи в порядке возрастания.)
\[-\frac{33\pi}{24}\]
\[-\frac{31\pi}{24}\]
\[-\frac{25\pi}{24}\]
\[-\frac{23\pi}{24}\]
Варианты ответов:
\[-\frac{65\pi}{12}\]
\[-\frac{31\pi}{24}\]
\[-\frac{77\pi}{12}\]
\[-\frac{25\pi}{24}\]
\[\frac{25\pi}{32}\]
\[\frac{17\pi}{32}\]
\[-\frac{33\pi}{24}\]
\[\frac{55\pi}{16}\]
\[\frac{47\pi}{16}\]
\[-\frac{23\pi}{24}\]
\[\frac{41\pi}{16}\]
\[-\frac{59\pi}{12}\]
\[\frac{49\pi}{16}\]
\[\frac{23\pi}{32}\]
\[-\frac{71\pi}{12}\]
\[\frac{31\pi}{32}\]