Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 884

\[\cos\left( 5x + \frac{\pi}{2} \right) + 2\sin x \bullet \cos{2x} = 0\]

\[- \sin{5x} + \sin( - x) + \sin{3x} = 0\]

\[\sin{3x} - \sin{5x} - \sin x = 0\]

\[2\sin( - x) \bullet \cos{4x} - \sin x = 0\]

\[- \sin x \bullet \left( 2\cos{4x} + 1 \right) = 0.\]

\[1)\ \sin x = 0\]

\[x = \pi n.\]

\[2)\ 2\cos{4x} + 1 = 0\]

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

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

\[4x = \pm \arccos\left( - \frac{1}{2} \right) + 2\pi n\]

\[4x = \pm \frac{2\pi}{3} + 2\pi n\]

\[x = \frac{1}{4}\left( \pm \frac{2\pi}{3} + 2\pi n \right)\]

\[x = \pm \frac{\pi}{6} + \frac{\text{πn}}{2}.\]

\[3)\left( - \frac{\pi}{6}\ \frac{\pi}{2} \right):\]

\[x_{наиб} = - \frac{\pi}{6} + \frac{\pi}{2} = - \frac{\pi}{6} + \frac{3\pi}{6} =\]

\[= \frac{2\pi}{6} = \frac{\pi}{3}.\]

\[Ответ:\ \ \frac{\pi}{3}.\]