Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1269

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\[\sin x\sin{2x} < \sin{3x}\sin{4x};\ \ \ \]

\[\left( 0;\frac{\pi}{2} \right).\]

\[f(x) = \sin x\sin{2x} -\]

\[- \sin{3x}\sin{4x} < 0\]

\[\frac{1}{2}\left( \cos x - \cos{3x} \right) =\]

\[= \frac{1}{2}\left( \cos x - \cos{7x} \right)\]

\[\cos{3x} - \cos{7x} = 0\]

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

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

\[5x = \text{πn}\]

\[x = \frac{\text{πn}}{5}.\]

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

\[2x = \text{πk}\]

\[x = \frac{\text{πk}}{2}.\]

\[0 < x < \frac{\pi}{5};\ \ \ \ \frac{2\pi}{5} < \pi < \frac{\pi}{2}.\]