Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 81

\[( - \pi;\ 2\pi).\]

\[1)\ tg\ x \geq 1\]

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

\[Ответ:\ - \frac{3\pi}{4} \leq x < - \frac{\pi}{2};\text{\ \ \ }\]

\[\frac{\pi}{4} \leq x < \frac{\pi}{2};\ \frac{5\pi}{4} \leq x < \frac{3\pi}{2}.\]

\[2)\ tg\ x < \frac{\sqrt{3}}{3}\]

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

\[Ответ:\ - \pi < x < - \frac{5\pi}{6};\]

\[- \frac{\pi}{2} < x < \frac{\pi}{6};\frac{\pi}{2} < x < \frac{7\pi}{6};\]

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

\[3)\ ctg\ x < - 1\]

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

\[Ответ:\ - \frac{\pi}{4} < x < 0;\]

\[\frac{3\pi}{4} < x < \pi;\frac{7\pi}{4} < x < 2\pi.\]

\[4)\ ctg\ x \geq - \sqrt{3}\]

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

\[Ответ:\ - \pi < x \leq - \frac{\pi}{6};\text{\ \ \ }\]

\[0 < x \leq \frac{5\pi}{6};\ \pi < x \leq \frac{11\pi}{6}.\]