ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1199

\[\boxed{\mathbf{1199}\mathbf{.}}\]

\[1)\ 4\sin^{4}x + \frac{1}{3}\cos^{2}x = \frac{1}{2}\]

\[4\sin^{4}x + \frac{1}{3} - \frac{1}{3}\sin^{2}x - \frac{1}{2} = 0\]

\[4\sin^{4}x - \frac{1}{3}\sin^{2}x - \frac{1}{6} = 0\ \ | \cdot 6\ \]

\[24\sin^{4}x - 2\sin^{2}x - 1 = 0\]

\[Пусть\sin^{2}x = t:\]

\[24t^{2} - 2t - 1 = 0\]

\[D_{1} = 1 + 24 = 25\]

\[t_{1} = \frac{1 + 5}{24} = \frac{1}{4};\ \ \ t_{2} = \frac{1 - 5}{24} =\]

\[= - \frac{1}{6} < 0\ (не\ подходит).\]

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

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

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

\[2x = \pm \frac{\pi}{3} + 2\pi k;\]

\[x = \pm \frac{\pi}{6} + \pi k.\]

\[Ответ:\ \ \pm \frac{\pi}{6} + \pi k.\]

\[2)\ 16\cos^{4}x + \sin^{2}x = \frac{7}{4}\]

\[16\cos^{4}x + 1 - \cos^{2}x - \frac{7}{4} = 0\]

\[16\cos^{4}x - \cos^{2}x - \frac{3}{4} = 0\ \ | \cdot 4\]

\[64\cos^{4}x - 4\cos^{2}x - 3 = 0\]

\[Пусть\cos^{2}x = t:\]

\[64t^{2} - 4t - 3 = 0\]

\[D_{1} = 4 + 192 = 196\]

\[t_{1} = \frac{2 - 14}{64} = - \frac{12}{64} =\]

\[= - \frac{3}{16} < 0\ (не\ подходит);\ \ \ \]

\[t_{2} = \frac{2 + 14}{64} = \frac{16}{64} = \frac{1}{4}.\]

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

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

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

\[2x = \pm \frac{2\pi}{3} + 2\pi k\]

\[x = \pm \frac{\pi}{3} + \pi k.\]

\[Ответ:\ \pm \frac{\pi}{3} + \pi k.\]