ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 725

\[\boxed{\mathbf{725}\mathbf{.}}\]

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

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

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

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

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

\[\frac{\pi}{6} < x_{1} < \frac{5\pi}{6};\]

\[\frac{13\pi}{6} < x_{2} < \frac{17\pi}{6}.\]

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

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

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

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

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

\[0 \leq x_{1} \leq \frac{\pi}{4};\]

\[\frac{3\pi}{4} \leq x_{2} \leq \frac{9\pi}{4};\]

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

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

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

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

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

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

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

\[0 \leq x_{1} \leq \frac{7\pi}{6};\]

\[\frac{11\pi}{6} \leq x_{2} \leq 3\pi.\]

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

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

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

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

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

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

\[\frac{4\pi}{3} < x_{1} < \frac{5\pi}{3}.\]