Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 633

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

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

\[\sin{4x} = \sin^{2}{4x}\]

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

\[4x = \arcsin 0 + \pi n = \pi n\]

\[x = \frac{1}{4} \bullet \pi n = \frac{\text{πn}}{4}.\]

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

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

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

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

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

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

\[1 - \sin^{4}x + \cos^{2}x = 0\]

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

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

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

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

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

\[\cos x = 0\]

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

\[\sin^{2}x + 2 = 0\]

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

\[корней\ нет.\]

\[Ответ:\ \ \frac{\pi}{2} + \pi n.\]