Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1116

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\[1)\cos{4x}\cos{2x} = \cos{5x}\cos x\]

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

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

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

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

\[Пусть\ t = \cos{2x}:\]

\[2t^{2} - t - 1 = 0\]

\[D = 1 + 8 = 9\]

\[t = \frac{1 - 3}{4} = - \frac{1}{2};\ \ \ t = \frac{1 + 3}{4} = 1.\]

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

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

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

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

\[2x = 2\pi k\]

\[x = \pi k.\]

\[2)\sin{5x}\sin x = \sin{7x}\sin{3x}\]

\[\frac{1}{2}\cos{4x} - \frac{1}{2}\cos{6x} = \frac{1}{2}\cos{4x} -\]

\[- \frac{1}{2}\cos{10x}\]

\[\cos{6x} - \cos{10x} = 0\]

\[\sin{8x}\sin{2x} = 0\]

\[1)\ \sin{8x} = 0\]

\[8x = \text{πk}\]

\[x = \frac{\text{πk}}{8}.\]

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

\[2x = \pi k\]

\[x = \frac{\text{πk}}{2}.\]

\[3)\sin\left( x + \frac{\pi}{3} \right)\cos\left( x - \frac{\pi}{6} \right) = 1\]

\[\frac{1}{2}\left( \sin\left( 2x + \frac{\pi}{6} \right) + \sin\frac{\pi}{2} \right) = 1\]

\[\sin\left( 2x + \frac{\pi}{6} \right) = 2 - 1\]

\[\sin\left( 2x + \frac{\pi}{6} \right) = 1\]

\[2x + \frac{\pi}{6} = \frac{\pi}{2} + 2\pi k\]

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

\[x = \frac{\pi}{6} + 2\pi k.\]

\[4)\ 2\sin\left( \frac{\pi}{4} + x \right)\sin\left( \frac{\pi}{4} - x \right) +\]

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

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

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

\[\cos x = 0\]

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