ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1211

\[\boxed{\mathbf{1211}\mathbf{.}}\]

\[1)\cos^{2}x + \cos^{2}{2x} = \cos^{2}{3x}\]

\[\cos^{2}x - \cos^{2}3x + \cos^{2}2x = 0\]

\[\left( \cos x + \cos{3x} \right) \cdot\]

\[\cdot \left( \cos x - \cos{3x} \right) + \cos^{2}2x = 0\]

\[2\cos x\cos{2x} \cdot 2\sin x\sin{2x} +\]

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

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

\[\cos{2x} \cdot\]

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

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

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

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

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

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

\[D = 1 + 16 = 17\]

\[\cos{2x} = \frac{1 + \sqrt{17}}{4} > 1 - не\ \]

\[имеет\ решений.\]

\[\cos{2x} = \frac{1 - \sqrt{17}}{4}\]

\[2x = \pm \frac{1 - \sqrt{17}}{4} + 2\pi k\]

\[x = \pm \frac{1}{2}\arccos\frac{1 - \sqrt{17}}{4} + \pi k.\]

\[Ответ:\ \ \frac{\pi}{4} + \frac{\text{πk}}{2};\]

\[\ \pm \frac{1}{2}\arccos\frac{1 - \sqrt{17}}{4} + \pi k.\text{\ \ }\]

\[2)\cos^{2}x + \cos^{2}{2x} +\]

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

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

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

\[\cos^{2}{3x} = \frac{1}{2}\left( 1 + \cos{6x} \right)\]

\[\frac{3}{2} +\]

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

\[= \frac{3}{2}\]

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

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

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

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

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

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

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

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

\[x = \frac{\pi}{8} + \frac{\text{πk}}{4}.\]

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

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

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

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

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

\[+ \cos^{2}{5x} = 2\]

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

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

\[+ \frac{3}{2}\cos{10x} = 2\]

\[\cos{4x} + \cos{6x} + \cos{8x} +\]

\[+ \cos{10x} = 0\]

\[2\cos{7x}\cos{3x} +\]

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

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

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

\[7x = \frac{\pi}{2} + \pi k\]

\[x = \frac{\pi}{14} + \frac{\text{πk}}{7}.\]

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

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

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

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

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

\[\cos x = 0\]

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

\[Ответ:\ \frac{\pi}{14} + \frac{\text{πk}}{7};\ \ \frac{\pi}{4} + \frac{\text{πk}}{2}.\]

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

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

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

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

\[\left( \sin x - \sin{4x} \right) \cdot\]

\[\cdot \left( \sin x + \sin{4x} \right) =\]

\[= \left( \sin{3x} - \sin{2x} \right) \cdot\]

\[\cdot \left( \sin{3x} + \sin{2x} \right)\]

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

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

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

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

\[4\cos\frac{5x}{2}\sin\frac{- 3x}{2}\sin\frac{5x}{2}\cos\frac{- 3x}{2} =\]

\[= 4\cos\frac{5x}{2}\sin\frac{x}{2}\sin\frac{5x}{2}\cos\frac{x}{2}\]

\[\sin{5x}\sin{- 3x} = \sin{5x}\sin x\]

\[\sin{5x}\left( \sin{( - 3x)} - \sin x \right) = 0\]

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

\[5x = \pi k\]

\[x = \frac{\text{πk}}{5}.\]

\[2\sin\frac{4x}{2}\cos\frac{2x}{2} = 0\]

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

\[2x = \pi k\]

\[x = \frac{\text{πk}}{2}.\]

\[\cos x = 0\]

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

\[Ответ:\ \frac{\text{πk}}{5};\ \ \frac{\pi}{2} + \pi k.\ \]