Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 73

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

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

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

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

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

\[\frac{\pi}{3} + 2\pi n < 2x < \frac{5\pi}{3} + 2\pi\text{n.}\]

\[Ответ:\ \ \frac{\pi}{6} + \pi n < x < \frac{5\pi}{6} + \pi\text{n.}\]

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

\[3\sin x - 2\left( 1 - \sin^{2}x \right) < 0\]

\[2\sin^{2}x + 3\sin x - 2 < 0\]

\[D = 9 + 16 = 25\]

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

\[\sin x_{2} = \frac{- 3 + 5}{2 \bullet 2} = \frac{1}{2}.\]

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

\[- 2 < \sin x < \frac{1}{2}.\]

\[Ответ:\ \]

\[- \frac{7\pi}{6} + 2\pi n < x < \frac{\pi}{6} + 2\pi n;\]