Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 672

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\[1)\cos^{3}x \bullet \sin x - \sin^{3}x \bullet \cos x = \frac{1}{4}\]

\[\sin x \bullet \cos x \bullet \left( \cos^{2}x - \sin^{2}x \right) = \frac{1}{4}\]

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

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

\[\sin{4x} = 1\]

\[4x = \arcsin 1 + 2\pi n\]

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

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

\[x = \frac{\pi}{8} + \frac{\text{πn}}{2}.\]

\[Ответ:\ \ \frac{\pi}{8} + \frac{\text{πn}}{2}.\]

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

\[\sin x \bullet \cos x \bullet \left( \sin^{2}x + \cos^{2}x \right) = \frac{1}{4}\]

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

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

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

\[2x = ( - 1)^{n} \bullet \arcsin\frac{1}{2} + \pi n\]

\[2x = ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n\]

\[x = \frac{1}{2} \bullet \left( ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n \right)\]

\[x = ( - 1)^{n} \bullet \frac{\pi}{12} + \frac{\text{πn}}{2}.\]

\[Ответ:\ \ ( - 1)^{n} \bullet \frac{\pi}{12} + \frac{\text{πn}}{2}.\]