ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 650

\[\boxed{\mathbf{650}\mathbf{.}}\]

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

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

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

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

\[Ответ:\ \ \]

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

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

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

\[x \leq \arcsin\frac{\sqrt{2}}{2} + 2\pi n\]

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

\[Ответ:\ \]

\[- \frac{5\pi}{4} + 2\pi n \leq x \leq \frac{\pi}{4} + 2\pi n.\]

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

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

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

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

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

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

\[Ответ:\ \ \]

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

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

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

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

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

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

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

\[Ответ:\ \]

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