Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 639

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

\[\sin x \bullet \sin{2x} \bullet \sin{3x} =\]

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

\[\sin x \bullet \sin{2x} \bullet \sin{3x} =\]

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

\[\sin x \bullet \frac{\cos{3x} + \cos{5x}}{2} = 0\]

\[\sin x \bullet \cos\frac{3x + 5x}{2} \bullet \cos\frac{3x - 5x}{2} = 0\]

\[\sin x \bullet \cos{4x} \bullet \cos x = 0\]

\[\frac{1}{2}\sin{2x} \bullet \cos{4x} = 0\]

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

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

\[x = \frac{1}{2} \bullet \pi n = \frac{\text{πn}}{2}.\]

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

\[4x = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n\]

\[x = \frac{1}{4} \bullet \left( \frac{\pi}{2} + \pi n \right) = \frac{\pi}{8} + \frac{\text{πn}}{4}.\]

\[Ответ:\ \ \frac{\text{πn}}{2};\ \ \frac{\pi}{8} + \frac{\text{πn}}{4}.\]

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

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

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

\[\left( \sin^{2}x - \cos^{2}x \right)^{2} = 0\]

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

\[- \cos{2x} = 0\]

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

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

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

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

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

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

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