Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1605

\[\boxed{\mathbf{1605}\mathbf{.}}\]

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

\[\cos^{2}{2x} \bullet 1^{2} + \frac{1}{8}\sin^{4}{2x} = a\]

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

\[\sin^{4}{2x} - 8\sin^{2}{2x} + (8 - 8a) = 0\]

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

\[= 64 - 32 + 32a = 32 + 32a =\]

\[= 16(2 + 2a)\]

\[При\ D \geq 0:\]

\[2 + 2a \geq 0\]

\[1 + a \geq 0\]

\[a \geq - 1.\]

\[\sin^{2}{2x} = \frac{8 \pm 4\sqrt{2 + 2a}}{2} =\]

\[= 4 \pm 2\sqrt{2 + 2a}\]

\[- 1 \leq \sin{2x} \leq 1\]

\[0 \leq \sin^{2}{2x} \leq 1.\]

\[1)\ 0 \leq 4 + 2\sqrt{2 + 2a} \leq 1\]

\[нет\ решений.\]

\[2)\ 0 \leq 4 - 2\sqrt{2 + 2a} \leq 1\]

\[- 4 \leq - 2\sqrt{2 + 2a} \leq - 3\]

\[3 \leq 2\sqrt{2 + 2a} \leq 4\]

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

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

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

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

\[\sin^{2}{2x} = 4 - \sqrt{2 + 2a}\]

\[\sin{2x} = \pm \sqrt{4 - \sqrt{2 + 2a}}\]

\[2x = \pm \arcsin\sqrt{4 - \sqrt{2 + 2a}} + \pi n\]

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

\[Ответ:\ \ \]