Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 442

\[\boxed{\mathbf{442.}}\]

\[1)\ a = \frac{\pi}{6}\]

\[0 < \frac{\pi}{6} < \frac{\pi}{2}\]

\[в\ I\ четверти.\]

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

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

\[\ во\text{\ II\ }четверти.\]

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

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

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

\[в\ III\ четверти.\]

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

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

\[в\ III\ четверти.\]

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

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

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

\[во\text{\ II\ }четверти.\]

\[6)\ a = 4,8\]

\[\pi \approx 3,14\]

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

\[в\ IV\ четверти.\]

\[7)\ a = - 1,31\]

\[\pi \approx 3,14\]

\[2\pi - 1,31 \approx 6,28 - 1,31 \approx 4,97\]

\[1,5\pi < 4,97 < 2\pi\]

\[в\ \text{IV\ }четверти.\]

\[8)\ a = - 2,7\]

\[\pi \approx 3,14\]

\[2\pi - 2,7 \approx 6,28 - 2,7 \approx 3,58\]

\[\pi < 3,58 < 1,5\pi\]

\[в\ III\ четверти.\]