Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 973

\[\boxed{\mathbf{973.}}\]

\[1)\ a = \frac{3\pi}{4}\]

\[\frac{\pi}{2} < \frac{3\pi}{4} < \pi\]

\[\ во\text{\ II\ }четверти.\]

\[2)\ a = - \frac{3\pi}{4}\]

\[2\pi - \frac{3\pi}{4} = \frac{8\pi}{4} - \frac{3\pi}{4} = \frac{5\pi}{4}\]

\[\pi < \frac{5\pi}{4} < \frac{3\pi}{2}\]

\[в\ III\ четверти.\]

\[3)\ a = - \frac{7\pi}{4}\]

\[2\pi - \frac{7\pi}{4} = \frac{8\pi}{4} - \frac{7\pi}{4} = \frac{\pi}{4}\]

\[\frac{\pi}{4} < \frac{3\pi}{4} < \pi\]

\[во\text{\ II\ }четверти.\]

\[4)\ a = \frac{7\pi}{6}\]

\[\pi < \frac{7\pi}{6} < \frac{3\pi}{2}\]

\[в\ III\ четверти.\]

\[5)\ a = - \frac{7\pi}{6}\]

\[2\pi - \frac{7\pi}{6} = \frac{12\pi}{6} - \frac{7\pi}{6} = \frac{5\pi}{6}\]

\[\frac{\pi}{2} < \frac{5\pi}{6} < \pi\]

\[во\text{\ II\ }четверти.\]

\[4)\ a = 4,8\]

\[\pi \approx 3,14\]

\[1,5\pi < 4,8 < 2\pi\]

\[в\ IV\ четверти.\]