Решебник по алгебре 11 класс Никольский Параграф 13. Использование свойств функции при решении уравнений и неравенств Задание 14

\[\boxed{\mathbf{14.}}\]

\[\textbf{а)}\ x^{2} - \pi x + \frac{\pi^{2}}{4} = \sin x - 1\]

\[\left( x - \frac{\pi}{2} \right)^{2} = \sin x - 1\]

\[x - \frac{\pi}{2} \geq 0;\]

\[\sin x - 1 \leq 0.\]

\[\left\{ \begin{matrix} \left( x - \frac{\pi}{2} \right)^{2} = 0 \\ \sin x - 1 = 0 \\ \end{matrix} \right.\ \]

\[x - \frac{\pi}{2} = 0\]

\[x = \frac{\pi}{2}.\]

\[Проверим:\]

\[\sin\frac{\pi}{2} - 1 = 1 - 1 = 0.\]

\[Ответ:x = \frac{\pi}{2}.\]

\[\textbf{б)}\ x^{2} - 4\pi x + 4\pi^{2} = \cos x - 1\]

\[(x - 2\pi)^{2} = \cos x - 1\]

\[(x - 2\pi)^{2} \geq 0;\]

\[\cos x - 1 \leq 0.\]

\[\left\{ \begin{matrix} (x - 2\pi)^{2} = 0 \\ \cos x - 1 = 0\ \\ \end{matrix} \right.\ \]

\[x - 2\pi = 0\]

\[x = 2\pi.\]

\[Проверим:\]

\[\cos{2\pi} - 1 = 1 - 1 = 0.\]

\[Ответ:x = 2\pi.\]

\[\textbf{в)}\ x^{2} + 2\pi x + \pi^{2} = \sin x - 1\]

\[(x + \pi)^{2} = \sin x - 1\]

\[(x + \pi)^{2} \geq 0;\]

\[\sin x - 1 \leq 0.\]

\[\left\{ \begin{matrix} (x + \pi)^{2} = 0 \\ \sin x - 1 = 0 \\ \end{matrix} \right.\ \]

\[(x + \pi)^{2} = 0\]

\[x + \pi = 0\]

\[x = - \pi.\]

\[Проверим:\]

\[\sin( - \pi) - 1 = 0 - 1 = - 1 \neq 0.\]

\[Не\ удовлетворяет.\]

\[Ответ:нет\ корней.\]

\[\textbf{г)}\ x^{2} - 2\pi x + \pi^{2} = \cos x - 1\]

\[(x - \pi)^{2} = \cos x - 1\]

\[(x - \pi)^{2} \geq 0;\]

\[\cos x - 1 \leq 0.\]

\[\left\{ \begin{matrix} (x - \pi)^{2} = 0\ \\ \cos x - 1 = 0 \\ \end{matrix} \right.\ \]

\[(x - \pi)^{2} = 0\]

\[x - \pi = 0\]

\[x = \pi.\]

\[Проверим:\]

\[\cos\pi - 1 = - 1 - 1 = - 2 \neq 0.\]

\[Не\ удовлетворяет.\]

\[Ответ:нет\ корней.\]