Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 878

\[1)\cos\sqrt{2 - x^{2}} = \frac{\sqrt{3}}{2}\]

\[0 \leq \sqrt{2 - x^{2}} \leq \sqrt{2}.\]

\[\sqrt{2 - x^{2}} = \arccos\frac{\sqrt{3}}{2}\]

\[\sqrt{2 - x^{2}} = \frac{\pi}{6}\]

\[2 - x^{2} = \frac{\pi^{2}}{36}\]

\[x^{2} = 2 - \frac{\pi^{2}}{36}\]

\[x = \pm \sqrt{2 - \frac{\pi^{2}}{36}}.\]

\[Ответ:\ \pm \sqrt{2 - \frac{\pi^{2}}{36}}.\]

\[2)\sin{\frac{5\pi}{4}x} = x^{2} - 4x + 5\]

\[x^{2} - 4x + 5 \geq - 1\]

\[x^{2} - 4x + 6 \geq 0\]

\[x \in R.\text{\ \ \ }\]

\[x^{2} - 4x + 5 \leq 1\]

\[x^{2} - 4x + 4 \leq 0\]

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

\[x = 2.\]

\[Проверка:\]

\[\sin\left( \frac{5\pi}{4} \bullet 2 \right) = \sin\frac{5\pi}{2} = \sin\frac{\pi}{2} = 1\]

\[2^{2} - 4 \bullet 2 + 5 = 4 - 8 + 5 = 1.\]

\(Ответ:\ \ 2.\)