Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 933

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

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

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

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

\[16\cos^{4}x - 2\cos^{2}x - 3 \leq 0\]

\[D = 4 + 192 = 196\]

\[\cos^{2}x_{1} = \frac{2 - 14}{2 \bullet 16} = - \frac{3}{8};\]

\[\cos^{2}x_{2} = \frac{2 + 14}{2 \bullet 16} = \frac{1}{2};\]

\[\left( \cos^{2}x + \frac{3}{8} \right)\left( \cos^{2}x - \frac{1}{2} \right) \leq 0\]

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

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

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

\[1)\ Верно\ при\ любом\ x:\]

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

\[\cos x \geq 0.\]

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

\[- \frac{3\pi}{4} + 2\pi n \leq x \leq \frac{3\pi}{4} + 2\pi n.\]

\[Ответ:\ \ \]

\[x \in \left( - \frac{3\pi}{4} + 2\pi n;\ \frac{3\pi}{4} + 2\pi n \right).\]