ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 885

\[\sin^{8}x + \cos^{8}x = a\]

\[1)\ y = \sin^{2}{2x}:\]

\[\left( 1 - \frac{1}{2}y \right)^{2} - \frac{1}{8}y^{2} = a\]

\[1 - y + \frac{1}{4}y^{2} - \frac{1}{8}y^{2} = a\]

\[\frac{1}{8}y^{2} - y + 1 - a = 0\]

\[y^{2} - 8y + 8(1 - a) = 0\]

\[D = 8^{2} - 4 \bullet 8(1 - a) =\]

\[= 64 - 32 + 32a = 32(1 + a)\]

\[y = \frac{8 \pm \sqrt{32(1 + a)}}{2} =\]

\[= \frac{8 \pm 4\sqrt{2(1 + a)}}{2} =\]

\[= 4 \pm 2\sqrt{2(1 + a)}.\]

\[2)\ Подставим:\]

\[\sin^{2}{2x} = 4 \pm 2\sqrt{2(1 + a)}\]

\[\frac{1 - \cos{4x}}{2} = 4 \pm 2\sqrt{2(1 + a)}\]

\[1 - \cos{4x} = 8 \pm 4\sqrt{2(1 + a)}\]

\[\cos{4x} = \mp 4\sqrt{2(1 + a)} - 7\]

\[\cos{4x} = 4\sqrt{2(1 + a)} - 7\]

\[4x = \pm \arccos\left( 4\sqrt{2(1 + a)} - 7 \right) + 2\pi n\]

\[x = \pm \frac{1}{4}\arccos\left( 4\sqrt{2(1 + a)} - 7 \right) + \frac{\text{πn}}{2}.\]

\[3)\ Область\ определения:\]

\[- 1 \leq 4\sqrt{2(1 + a)} - 7 \leq 1\]

\[6 \leq 4\sqrt{2 + 2a} \leq 8\]

\[\frac{3}{2} \leq \sqrt{2 + 2a} \leq 2\]

\[\frac{9}{4} \leq 2 + 2a \leq 4\]

\[\frac{1}{4} \leq 2a \leq 2\]

\[\frac{1}{8} \leq a \leq 1.\]

\[Ответ:\ \]

\[\pm \frac{1}{4}\arccos\left( 4\sqrt{2(1 + a)} - 7 \right) + \frac{\text{πn}}{2};\text{\ \ \ }\]

\[\frac{1}{8} \leq a \leq 1.\]