Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 640

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\[\sin{2x} \bullet \left( \sin{4x} + \sin{6x} \right) = 0\]

\[\sin{2x} \bullet 2 \bullet \sin\frac{4x + 6x}{2} \bullet \cos\frac{4x - 6x}{2} = 0\]

\[2 \bullet \sin{2x} \bullet \sin\frac{10x}{2} \bullet \cos\left( - \frac{2x}{2} \right) = 0\]

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

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

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

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

\[2)\ \sin{5x} = 0\]

\[5x = \arcsin 0 + \pi n = \pi n\]

\[x = \frac{1}{5} \bullet \pi n = \frac{\text{πn}}{5}.\]

\[3)\ \cos x = 0\]

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

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

\[Ответ:\ \ \frac{\text{πn}}{2};\ \ \frac{\text{πn}}{5}.\]

\[2)\sin^{6}x + \cos^{6}x = \frac{1}{4}\]

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

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

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

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

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

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

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

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

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

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