Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1221

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\[1)\cos x \geq \frac{\sqrt{2}}{2}\]

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

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

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

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

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

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

\[- \arccos\frac{\sqrt{3}}{2} + 2\pi n\]

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

\[Ответ:\ \ \frac{\pi}{6} + 2\pi n < x < \frac{11\pi}{6} +\]

\[+ 2\pi n.\]

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

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

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

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

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

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

\[+ 2\pi n\]

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

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

\[4)\cos x \leq - \frac{\sqrt{2}}{2}\ \]

\[\arccos\left( - \frac{\sqrt{2}}{2} \right) +\]

\[+ 2\pi n \leq x \leq 2\pi -\]

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

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

\[- \pi + \arccos\frac{\sqrt{2}}{2} + 2\pi n\]

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

\[Ответ:\ \ \frac{3\pi}{4} + 2\pi n \leq x \leq \frac{5\pi}{4} +\]

\[+ 2\pi n.\]