Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 872

\[1)\cos x \bullet \sin{9x} = \cos{3x} \bullet \sin{7x}\]

\[\frac{1}{2}\left( \sin(9x - x) + \sin(9x + x) \right) =\]

\[= \frac{1}{2}\left( \sin(7x - 3x) + \sin(7x + 3x) \right)\]

\[\sin{8x} + \sin{10x} = \sin{4x} + \sin{10x}\]

\[\sin{8x} - \sin{4x} = 0\]

\[2 \bullet \cos\frac{8x + 4x}{2} \bullet \sin\frac{8x - 4x}{2} = 0\]

\[\cos{6x} \bullet \sin{2x} = 0\]

\[1)\ \cos{6x} = 0\]

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

\[x = \frac{\pi}{12} + \frac{\text{πn}}{6}.\]

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

\[2x = \pi n\]

\[x = \frac{\text{πn}}{2}.\]

\[Ответ:\ \ \frac{\pi}{12} + \frac{\text{πn}}{6};\ \frac{\text{πn}}{2}.\]

\[2)\sin x \bullet \cos{5x} = \sin{9x} \bullet \cos{3x}\]

\[\frac{1}{2}\left( \sin(x - 5x) + \sin(x + 5x) \right) =\]

\[= \frac{1}{2}\left( \sin(9x - 3x) + \sin(9x + 3x) \right)\]

\[\sin( - 4x) + \sin{6x} = \sin{6x} + \sin{12x}\]

\[\sin{12x} + \sin{4x} = 0\]

\[2 \bullet \sin\frac{12x + 4x}{2} \bullet \cos\frac{12x - 4x}{2} = 0\]

\[\sin{8x} \bullet \cos{4x} = 0\]

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

\[8x = \pi n\]

\[x = \frac{\text{πn}}{8}.\]

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

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

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

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