Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 688

\[\boxed{\mathbf{688}\mathbf{.}}\]

\[\sin^{10}x + \cos^{10}x = a\]

\[\left( \sin^{2}x \right)^{5} + \left( \cos^{2}x \right)^{5} = a\]

\[\frac{\left( 1 - \cos{2x} \right)^{5}}{2^{5}} + \frac{\left( 1 + \cos{2x} \right)^{5}}{2^{5}} = a\]

\[\frac{2 + 20\cos^{2}{2x} + 10\cos^{4}{2x}}{32} = a\]

\[10\cos^{4}{2x} + 20\cos^{2}{2x} + 2 - 32a = 0\]

\[5\cos^{4}{2x} + 10\cos^{2}{2x} + (1 - 16a) = 0\]

\[y = \cos^{2}{2x}:\]

\[5y^{2} + 10y + (1 - 16a) = 0\]

\[D = 10^{2} - 4 \bullet 5 \bullet (1 - 16a) =\]

\[= 100 - 20 + 320a = 320a + 80\]

\[y = \frac{- 10 \pm \sqrt{320a + 80}}{2 \bullet 5} =\]

\[= - 1 \pm \frac{\sqrt{320a + 80}}{10};\]

\[y = - 1 - \frac{\sqrt{320a + 80}}{10} < 1 -\]

\[не\ подходит.\]

\[Имеет\ корни\ при:\]

\[- 1 \leq \cos^{2}{2x} \leq 1\]

\[0 \leq \cos{2x} \leq 1\]

\[0 \leq - 1 + \frac{\sqrt{320a + 80}}{10} \leq 1\]

\[1 \leq \frac{\sqrt{320a + 80}}{10} \leq 2\]

\[10 \leq \sqrt{320a + 80} \leq 20\]

\[100 \leq 320a + 80 \leq 400\]

\[20 \leq 320a \leq 320\]

\[\frac{1}{16} \leq a \leq 1.\]

\[Ответ:\ \ \frac{1}{16} \leq a \leq 1.\]