ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 682

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\[\cos^{2}x + \cos^{2}{2x} + \cos^{2}{3x} = \frac{3}{2}\]

\[\cos^{2}x + \cos^{2}{2x} + \cos^{2}{3x} =\]

\[\frac{1}{2}\left( \cos{2x} + \cos{4x} + \cos{6x} \right) = 0\]

\[\cos{2x} + \cos{6x} + \cos{4x} = 0\]

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

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

\[\cos{4x} \bullet \left( 2\cos{2x} + 1 \right) = 0\]

\[1)\ \cos{4x} = 0\]

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

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

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

\[2)\ 2\cos{2x} + 1 = 0\]

\[2\cos{2x} = - 1\]

\[\cos{2x} = - \frac{1}{2}\]

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

\[2x = \pm \left( \pi - \frac{\pi}{3} \right) + 2\pi n\]

\[2x = \pm \frac{2\pi}{3} + 2\pi n\]

\[x = \frac{1}{2} \bullet \left( \pm \frac{2\pi}{3} + 2\pi n \right)\]

\[x = \pm \frac{\pi}{3} + \pi n.\]

\[Ответ:\ \ \frac{\pi}{8} + \frac{\text{πn}}{4};\ \ \pm \frac{\pi}{3} + \pi n.\]