Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1212

\[\boxed{\mathbf{1212}\mathbf{.}}\]

\[1)\cos{7x}\cos{13x} = \cos x\cos{19x}\]

\[\frac{1}{2}\left( \cos{20x} + \cos{6x} \right) =\]

\[= \frac{1}{2}\left( \cos{20x} + \cos{18x} \right)\]

\[\cos{6x} - \cos{18x} = 0\]

\[\sin{12x}\sin{6x} = 0\]

\[\sin{12x} = 0\]

\[12x = \pi n\]

\[x = \frac{\text{πn}}{12}.\]

\[\sin{6x} = 0\]

\[6x = \pi n\]

\[x = \frac{\text{πn}}{6}.\]

\[Ответ:x = \frac{\text{πn}}{12}.\]

\[2)\sin x\sin{5x} = \sin{2x}\sin{4x}\]

\[\frac{\cos(x - 5x) - \cos(x + 5x)}{2} =\]

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

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

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

\[\cos{4x} - \cos{2x} = 0\]

\[2\sin\frac{6x}{2}\sin\frac{2x}{2} = 0\]

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

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

\[3x = \pi n\]

\[x = \frac{\text{πn}}{3}.\]

\[\sin x = 0\]

\[x = \pi n.\]

\[Ответ:\ \ x = \frac{\text{πn}}{3}.\]

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

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

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

\[Пусть\cos{2x} = y:\]

\[2y^{2} + y - 2 = 0\]

\[D = 1 + 16 = 17\]

\[y = \frac{- 1 - \sqrt{17}}{4} - не\ подходит.\]

\[y = \frac{\sqrt{17} - 1}{4}\]

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

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

\[Ответ:\ \ \pm \frac{1}{2}\arccos\frac{\sqrt{17} - 1}{4} +\]

\[+ \text{πn.}\]

\[4)\sin x\sin{3x} = \frac{1}{2}\]

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

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

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

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

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

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

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

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

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

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

\[x = \pm \frac{\pi}{6} + \pi n.\]

\[Ответ:x = \frac{\pi}{4} + \frac{\text{πn}}{2};\ \]

\[\ x = = \pm \frac{\pi}{6} + \pi n.\]