Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1565

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\[\frac{\sin{3x}}{\sin x} - \frac{\sin x}{\sin{3x}} = 2\cos{2x}\]

\[\frac{\sin^{3}{3x} - \sin^{2}x}{\sin x \bullet \sin{3x}} - 2\cos{2x} = 0\]

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

\[2x = \arccos 0 + \pi n\]

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

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

\[x = \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

\[2)\ \sin x = 0\]

\[x = \arcsin 0 + \pi n = \pi n.\]

\[3)\ 4\sin x \bullet \cos^{2}x - \sin{3x} = 0\]

\[\sin x = 0\]

\[x = \arcsin 0 + \pi n = \pi n.\]

\[Имеет\ смысл\ при:\]

\[\sin x \neq 0\]

\[x \neq \arcsin 0 + \pi n \neq \pi n.\]

\[\sin{3x} \neq 0\]

\[3x \neq \arcsin 0 + \pi n \neq \pi n\]

\[x \neq \frac{\text{πn}}{3}.\]

\[Ответ:\ \ \frac{\pi}{4} + \frac{\text{πn}}{2}.\]