ГДЗ по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 48

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\[\textbf{а)}\ \left( \sin x - \frac{1}{2} \right)\left( \sin x + \frac{\sqrt{2}}{2} \right) > 0\]

\[M = (0;\ \pi).\]

\[1)\ \left\{ \begin{matrix} \sin x - \frac{1}{2} > 0\ \ \\ \sin x + \frac{\sqrt{2}}{2} > 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \sin x > \frac{1}{2}\text{\ \ \ \ \ \ } \\ \sin x > - \frac{\sqrt{2}}{2} \\ \end{matrix} \right.\ \]

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

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

\[2)\ \left\{ \begin{matrix} \sin x - \frac{1}{2} < 0\ \ \ \\ \sin x + \frac{\sqrt{2}}{2} < 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \sin x < \frac{1}{2}\text{\ \ \ \ \ \ } \\ \sin x < - \frac{\sqrt{2}}{2} \\ \end{matrix} \right.\ \]

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

\[- \frac{3\pi}{4} + 2\pi k < x < - \frac{\pi}{4} + 2\pi k.\]

\[Решение\ неравенства:\]

\[x \in \left( \frac{\pi}{6};\frac{5\pi}{6} \right).\]

\[Ответ:x \in \left( \frac{\pi}{6};\frac{5\pi}{6} \right).\]

\[\textbf{б)}\ \left( \cos x + \frac{1}{2} \right)\left( \cos x - \frac{\sqrt{3}}{2} \right) > 0\]

\[M = \left( - \frac{\pi}{2};\frac{\pi}{2} \right).\]

\[1)\ \left\{ \begin{matrix} \cos x + \frac{1}{2} > 0\ \ \\ \cos x - \frac{\sqrt{3}}{2} > 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \cos x > - \frac{1}{2} \\ \cos x > \frac{\sqrt{3}}{2}\ \\ \end{matrix} \right.\ \]

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

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

\[2)\ \left\{ \begin{matrix} \cos x + \frac{1}{2} < 0\ \ \\ \cos x - \frac{\sqrt{3}}{2} < 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \cos x < - \frac{1}{2} \\ \cos x < \frac{\sqrt{3}}{2}\ \\ \end{matrix} \right.\ \]

\[\cos x < - \frac{1}{2}\]

\[\frac{2\pi}{3} + 2\pi k < x < \frac{4\pi}{3} + 2\pi k.\]

\[Решение\ неравенства:\]

\[x \in \left( - \frac{\pi}{6};\frac{\pi}{6} \right).\]

\[Ответ:x \in \left( - \frac{\pi}{6};\frac{\pi}{6} \right).\]