Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 50

\[1)\cos^{2}x > \frac{3}{4}\]

\[\frac{1 + \cos{2x}}{2} > \frac{3}{4}\]

\[2 + 2\cos{2x} > 3\]

\[2\cos{2x} > 1\]

\[\cos{2x} > \frac{1}{2}\]

\[- \frac{\pi}{3} + 2\pi n < 2x < \frac{\pi}{3} + 2\pi\text{n.}\]

\[Ответ:\ \]

\[- \frac{\pi}{6} + \pi n < x < \frac{\pi}{6} + \pi\text{n.}\]

\[2)\ 2\cos^{2}x - 3\cos x - 2 > 0\]

\[D = 9 + 16 = 25\]

\[\cos x_{1} = \frac{3 - 5}{2 \bullet 2} = - \frac{1}{2};\]

\[\cos x_{2} = \frac{3 + 5}{2 \bullet 2} = 2.\]

\[\left( \cos x + \frac{1}{2} \right)\left( \cos x - 2 \right) > 0\]

\[\cos x < - \frac{1}{2}\]

\[\cos x > 2.\]

\[Ответ:\ \ \]

\[\frac{2\pi}{3} + 2\pi n < x < \frac{4\pi}{3} + 2\pi\text{n.}\]

\[3)\ \sqrt[4]{\frac{7 - \cos{4x}}{2}} > - 2\cos x\]

\[\sqrt[4]{3 + \frac{1 - \cos{4x}}{2}} > - 2\cos x\]

\[\sqrt[4]{3 + \sin^{2}{2x}} > - 2\cos x\]

\[\sqrt[4]{3 + 4\sin^{2}x\cos^{2}x} > - 2\cos x\]

\[\sqrt[4]{3 + 4\cos^{2}x\left( 1 - \cos^{2}x \right)} > - 2\cos x\]

\[\sqrt[4]{3 + 4\cos^{2}x - 4\cos^{4}x} > - 2\cos x\]

\[3 + 4\cos^{2}x - 4\cos^{4}x > 16\cos^{4}x\]

\[20\cos^{4}x - 4\cos^{2}x - 3 < 0\]

\[D = 16 + 240 = 256\]

\[\cos^{2}x_{1} = \frac{4 - 16}{2 \bullet 20} = - \frac{3}{10};\ \]

\[\cos^{2}x_{2} = \frac{4 + 16}{2 \bullet 20} = \frac{1}{2}.\]

\[\left( \cos^{2}x + \frac{3}{10} \right)\left( \cos^{2}x - \frac{1}{2} \right) < 0\]

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

\[\frac{1 + \cos{2x}}{2} - \frac{1}{2} < 0\]

\[\cos{2x} < 0\]

\[\frac{\pi}{2} + 2\pi n < 2x < \frac{3\pi}{2} + 2\pi n\]

\[\frac{\pi}{4} + \pi n < x < \frac{3\pi}{4} + \pi\text{n.}\]

\[Всегда\ верно:\]

\[- 2\cos x < 0\]

\[\cos x > 0\]

\[- \frac{\pi}{2} + 2\pi n < x < \frac{\pi}{2} + 2\pi n.\]

\[Ответ:\ \]

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