ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1228

\[{\boxed{\mathbf{1228}\mathbf{.}} }{1)\ \frac{\sin^{2}x - \frac{1}{4}}{\sqrt{3} - \left( \sin x + \cos x \right)} > 0}\]

\[\left\{ \begin{matrix} \sin^{2}x - \frac{1}{4} > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \sqrt{3} - \left( \sin x + \cos x \right) > 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }или\ \ \ \ \]

\[\left\{ \begin{matrix} \sin^{2}x - \frac{1}{4} < 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \sqrt{3} - \left( \sin x + \cos x \right) < 0 \\ \end{matrix} \right.\ \]

\[\sin^{2}x - \frac{1}{4} = \frac{1 - \cos{2x}}{2} - \frac{1}{4} =\]

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

\[\frac{1}{4} - \frac{1}{2}\cos{2x} > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \]

\[\frac{1}{4} - \frac{1}{2}\cos{2x} < 0\ \]

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

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

\[\sqrt{3} - \left( \sin x + \cos x \right) = \sqrt{3} -\]

\[- \sqrt{2}\sin\left( x + \frac{\pi}{4} \right)\]

\[\sqrt{3} - \sqrt{2}\sin\left( x + \frac{\pi}{4} \right) > 0\]

\[\sin\left( x + \frac{\pi}{4} \right) < \frac{\sqrt{6}}{2} \rightarrow x - любое\]

\[\ число;\]

\[\sqrt{3} - \sqrt{2}\sin\left( x + \frac{\pi}{4} \right) < 0\]

\[\sin\left( x + \frac{\pi}{4} \right) > \frac{\sqrt{6}}{2} \rightarrow нет\]

\[\ решений.\]

\[Получаем:\]

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

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

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

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

\[\left\{ \begin{matrix} - 2\cos x < 0\ \ \ \\ \frac{7 - \cos{4x}}{2} \geq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }и\ \ \ \ \ \ \ \]

\[\left\{ \begin{matrix} - 2\cos x \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{7 - \cos{4x}}{2} > 16\cos^{4}x \\ \end{matrix} \right.\ \]

\[1)\cos x > 0;\ \cos{4x} < 7\]

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

\[2)\ t = \pm \frac{\sqrt{2}}{2}\]

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

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

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

\[\left\{ \begin{matrix} - 2\sin x < 0\ \ \\ \frac{7 - \cos{4x}}{2} \geq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }и\ \ \ \ \ \]

\[\left\{ \begin{matrix} - 2\sin x \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{7 - \cos{4x}}{2} > 16\sin^{4}x \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \sin x > 0\ \\ \sin{4x} < 7 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} \sin x \leq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - 20\cos^{4}x + 36\cos^{2}x - 13 > 0 \\ \end{matrix} \right.\ \]

\[1)\sin x > 0\]

\[0 < x < \pi.\]

\[2)\ 20t^{2} - 36t + 13 < 0\]

\[D_{1} = 324 - 260 = 64\]

\[t_{1} = \frac{18 + 8}{20} = \frac{26}{20} = \frac{13}{10} = 1,3;\ \]

\[\ t_{2} = \frac{18 - 8}{20} = \frac{1}{2}.\]

\[0,5 < t < 1,3.\]

\[\frac{1}{2} < \cos^{2}x < 1,3\]

\[- \frac{\sqrt{130}}{10} < \cos x < - \frac{\sqrt{2}}{2}\]

\[\frac{\sqrt{2}}{2} < \cos x < \frac{\sqrt{130}}{10}\]

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