Решебник по алгебре 11 класс Никольский Параграф 13. Использование свойств функции при решении уравнений и неравенств Задание 23

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\[\textbf{а)}\ tg^{2}x + ctg^{2}x =\]

\[= 2\sin^{2}\sqrt{\frac{5\pi^{2}}{16} - x^{2}}\]

\[tg^{2}x + \frac{1}{tg^{2}x} = 2\sin^{2}\sqrt{\frac{5\pi^{2}}{16} - x^{2}}\]

\[tg^{2}x + \frac{1}{tg^{2}x} \geq 2;\]

\[2\sin^{2}\sqrt{\frac{5\pi^{2}}{16} - x^{2}} \leq 2;\]

\[\left\{ \begin{matrix} tg^{2}x + \frac{1}{tg^{2}x} = 2\ \ \ \ \ \ \ \ \ \\ 2\sin^{2}\sqrt{\frac{5\pi^{2}}{16} - x^{2}} = 2 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} tg^{2}x = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \sin^{2}\sqrt{\frac{5\pi^{2}}{16} - x^{2}} = 1 \\ \end{matrix} \right.\ \]

\[tg^{2}x = 1\]

\[tgx = \pm 1\]

\[x = \pm \frac{\pi}{4} + \pi k.\]

\[\sin^{2}\sqrt{\frac{5\pi^{2}}{16} - x^{2}} = 1\]

\[Проверим\ x = - \frac{\pi}{4};n = 0:\]

\[Проверим\ x = \frac{\pi}{4};\ \ n = 0:\]

\[Ответ:x = \pm \frac{\pi}{4}.\]

\[\textbf{б)}\ \frac{\cos^{2}{2x}}{\sin^{8}x} + \frac{\sin^{8}x}{\cos^{2}{2x}} =\]

\[= 2\cos^{2}\sqrt{\frac{\pi^{2}}{4} - x^{2}}\]

\[\frac{\cos^{2}{2x}}{\sin^{8}x} + \frac{1}{\frac{\cos^{2}{2x}}{\sin^{8}x}} =\]

\[= 2\cos^{2}\sqrt{\frac{\pi^{2}}{4} - x^{2}}\]

\[\frac{\cos^{2}{2x}}{\sin^{8}x} + \frac{1}{\frac{\cos^{2}{2x}}{\sin^{8}x}} \geq 2;\]

\[2\cos^{2}\sqrt{\frac{\pi^{2}}{4} - x^{2}} \leq 2;\]

\[\left\{ \begin{matrix} \frac{\cos^{2}{2x}}{\sin^{8}x} + \frac{1}{\frac{\cos^{2}{2x}}{\sin^{8}x}} = 2 \\ 2\cos^{2}\sqrt{\frac{\pi^{2}}{4} - x^{2}} = 2\ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \frac{\cos^{2}{2x}}{\sin^{8}x} = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \cos^{2}\sqrt{\frac{\pi^{2}}{4} - x^{2}} = 1 \\ \end{matrix} \right.\ \]

\[\cos^{2}{2x} = \sin^{8}x.\]

\[\cos^{2}\sqrt{\frac{\pi^{2}}{4} - x^{2}} = 1\]

\[Проверим\ x = - \frac{\pi}{2}:\]

\[Проверим\ x = \frac{\pi}{2}:\]

\[Ответ:x = \pm \frac{\pi}{2}.\]