ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1256

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

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

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

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

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

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

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

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

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

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

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

\[\sin^{6}x + \cos^{6}x =\]

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

\[\bullet \left( \sin^{4}x - \sin^{2}x\cos^{2}x + \cos^{4}x \right) =\]

\[= \left( \sin^{2}x + \cos^{2}x \right) - \frac{3}{4}\sin^{2}{2x} =\]

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

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

\[1 - \frac{3}{4} \cdot \frac{1 - \cos{4x}}{2} = \frac{1}{4}\]

\[\cos{4x} = - 1\]

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

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

\[3)\sin^{2}x + \sin^{2}{2x} = \sin^{2}{3x}\]

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

\[= \sin^{2}{3x}\]

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

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

\[\cos{3x}\left( \cos x - \cos{3x} \right) = 0\]

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

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

\[x = \frac{\pi}{6} + \frac{\text{πn}}{3}.\]

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

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

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

\[2x = \pi n\]

\[x = \frac{\text{πn}}{2}.\]

\[\sin x = 0\]

\[x = \pi n.\]

\[4)\cos^{2}x + \cos^{2}{2x} = \sin^{2}{3x} +\]

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

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

\[= \frac{1}{2} - \frac{1}{2}\cos{6x} + \frac{1}{2} - \frac{1}{2}\cos{8x}\]

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

\[= - \left( \cos{6x} + \cos{8x} \right)\]

\[2\cos{3x}\cos x = - 2\cos{7x}\cos x\]

\[2\cos x(\cos{3x + \cos{7x})} = 0\]

\[\cos x = 0\]

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

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

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

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

\[5x = \frac{\pi}{2} + \pi n\]

\[x = \frac{\pi}{10} + \frac{\text{πn}}{5}.\]

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

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

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