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МатематикаГДЗ по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 16
Задание 16
\[\boxed{\mathbf{16.}}\]
\[\textbf{а)}\ 1 + \sin x > |\cos x|\]
\[\left( 1 + \sin x \right)^{2} > cos^{2}x\]
\[1 + 2\sin x + sin^{2}x > 1 - sin^{2}\text{x\ }\]
\[2sin^{2}x + 2\sin x > 0\]
\[2\sin x\left( \sin x + 1 \right) > 0\]
\[1)\ \left\{ \begin{matrix} \sin x > 0\ \ \ \ \ \ \ \ \\ \sin x + 1 > 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} \sin x > 0\ \ \ \\ \sin x > - 1 \\ \end{matrix} \right.\ \]
\[x \in (2\pi k;\ \pi + 2\pi k).\]
\[2)\ \left\{ \begin{matrix} \sin x < 0\ \ \ \ \ \ \ \ \\ \sin x + 1 < 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} \sin x < 0\ \ \ \ \ \ \ \\ \sin x < - 1\ \ \ \\ \end{matrix} \right.\ \]
\[нет\ решений.\]
\[Ответ:\ x \in (2\pi k;\ \pi + 2\pi k).\]
\[\textbf{б)}\ 1 - \cos x > |\sin x|\]
\[\left( 1 - \cos x \right)^{2} > sin^{2}x\]
\[1 - 2\cos x + cos^{2}x > 1 - cos^{2}x\]
\[2cos^{2}x - 2\cos x > 0\]
\[2\cos x\left( \cos x - 1 \right) > 0\]
\[1)\ \left\{ \begin{matrix} \cos x > 0\ \ \ \ \ \ \ \ \\ \cos x - 1 > 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} \cos x > 0 \\ \cos x > 1 \\ \end{matrix} \right.\ \]
\[нет\ решений.\]
\[2)\ \left\{ \begin{matrix} \cos x < 0\ \ \ \ \ \ \ \ \\ \cos x - 1 < 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} \cos x < 0 \\ \cos x < 1 \\ \end{matrix} \right.\ \]
\[x \in \left( 2\pi k + \frac{\pi}{2};\frac{3\pi}{2} + 2\pi k \right).\]
\[Ответ:\ x \in \left( 2\pi k + \frac{\pi}{2};\frac{3\pi}{2} + 2\pi k \right).\]
\[\textbf{в)}\ 1 - \sin x < |\cos x|\]
\[\left( 1 - \sin x \right)^{2} < cos^{2}x\]
\[1 - 2\sin x + sin^{2}x < 1 - sin^{2}x\]
\[2sin^{2}x - 2\sin x < 0\]
\[2\sin x\left( \sin x - 1 \right) < 0\]
\[1)\ \left\{ \begin{matrix} \sin x < 0\ \ \ \ \ \ \ \ \\ \sin x - 1 > 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} \sin x < 0 \\ \sin x > 1 \\ \end{matrix} \right.\ \]
\[нет\ решений.\]
\[2)\ \left\{ \begin{matrix} \sin x > 0\ \ \ \ \ \ \ \ \\ \sin x - 1 < 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} \sin x > 0 \\ \sin x < 1 \\ \end{matrix} \right.\ \]
\[x \in (2\pi k;\ \pi + 2\pi k).\]
\[Ответ:\ x \in (2\pi k;\ \pi + 2\pi k)\text{.\ \ }\]
\[\textbf{г)}\ 1 + \cos x < \left| \sin x \right|\]
\[\left( 1 + \cos x \right)^{2} < sin^{2}x\]
\[1 + 2\cos x + cos^{2}x < 1 - cos^{2}x\]
\[2\text{co}s^{2}x + 2\cos x < 0\]
\[2\cos x\left( \cos x + 1 \right) < 0\]
\[1)\ \left\{ \begin{matrix} \cos x < 0\ \ \ \ \ \ \ \ \\ \cos x + 1 > 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} \cos x < 0\ \ \ \\ \cos x > - 1 \\ \end{matrix} \right.\ \]
\[x \in \left( 2\pi k + \frac{\pi}{2};\frac{3\pi}{2} + 2\pi k \right).\]
\[2)\ \left\{ \begin{matrix} \cos x > 0\ \ \ \ \ \ \ \ \\ \cos x + 1 < 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} \cos x > 0\ \ \ \\ \cos x < - 1 \\ \end{matrix} \right.\ \]
\[нет\ решений.\]
\[Ответ:\ x \in \left( 2\pi k + \frac{\pi}{2};\frac{3\pi}{2} + 2\pi k \right).\]