ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 934

\[\sqrt[4]{\frac{7 - \cos{4x}}{2}} > - 2\sin x\]

\[\frac{8}{2} - \frac{1 + \cos{4x}}{2} > 16\sin^{4}x\]

\[4 + \cos^{2}{2x} \geq 16\sin^{4}x\]

\[4 + 1 - 2\sin^{2}x \geq 16\sin^{4}x\]

\[16\sin^{4}x + 2\sin^{2}x - 5 \leq 0\]

\[D = 4 + 320 = 324\]

\[\sin^{2}x_{1} = \frac{- 2 - 18}{2 \bullet 16} = - \frac{5}{8};\]

\[\sin^{2}x_{2} = \frac{- 2 + 18}{2 \bullet 16} = \frac{1}{2};\]

\[\left( \sin^{2}x + \frac{5}{8} \right)\left( \sin^{2}x - \frac{1}{2} \right) \leq 0\]

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

\[\left( \sin x + \frac{\sqrt{2}}{2} \right)\left( \sin x - \frac{\sqrt{2}}{2} \right) \leq 0\]

\[- \frac{\sqrt{2}}{2} \leq \sin x \leq \frac{\sqrt{2}}{2}.\]

\[1)\ Верно\ при\ любом\ x:\]

\[- 2\sin x \leq 0\]

\[\sin x \geq 0.\]

\[2)\ \sin x \geq - \frac{\sqrt{2}}{2}\]

\[- \frac{\pi}{4} + 2\pi n \leq x \leq \frac{5\pi}{4} + 2\pi n.\]

\[Ответ:\ \ \]

\[x \in \left( - \frac{\pi}{4} + 2\pi n;\ \frac{5\pi}{4} + 2\pi n \right).\]