Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1225

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\[1)\ \sqrt{2}\cos{2x} \leq 1\]

\[\cos{2x} \leq \frac{1}{\sqrt{2}}\]

\[\arccos\frac{1}{\sqrt{2}} + 2\pi n \leq 2x \leq 2\pi -\]

\[- \arccos\frac{1}{\sqrt{2}} + 2\pi n\]

\[\frac{\pi}{4} + 2\pi n \leq 2x \leq 2\pi - \frac{\pi}{4} + 2\pi n\]

\[\frac{\pi}{4} + 2\pi n \leq 2x \leq \frac{7\pi}{4} + 2\pi n\]

\[Ответ:\ \frac{\pi}{8} + \pi n \leq x \leq \frac{7\pi}{8} + \pi n.\]

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

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

\[\arcsin\left( - \frac{1}{2} \right) + 2\pi n < 3x < \pi -\]

\[- \arcsin\left( - \frac{1}{2} \right) + 2\pi n\]

\[- \arcsin\frac{1}{2} + 2\pi n < 3x < \pi +\]

\[+ \arcsin\frac{1}{2} + 2\pi n\]

\[- \frac{\pi}{6} + 2\pi n < 3x < \pi + \frac{\pi}{6} + 2\pi n\]

\[- \frac{\pi}{6} + 2\pi n < 3x < \frac{7\pi}{6} + 2\pi n\]

\[Ответ:\ - \frac{\pi}{18} + \frac{2\pi n}{3} < x < \frac{7\pi}{18} +\]

\[+ \frac{2\pi n}{3}.\]