Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1240

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\[1)\cos^{2}{3x} - \cos{3x} \bullet \cos{5x} = 0\]

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

\[\cos{3x} \bullet ( - 2) \bullet \sin\frac{3x + 5x}{2} \bullet\]

\[\bullet \sin\frac{3x - 5x}{2} = 0\]

\[- 2 \bullet \cos{3x} \bullet \sin\frac{8x}{2} \bullet\]

\[\bullet \sin\left( - \frac{2x}{2} \right) = 0\]

\[2 \bullet \cos{3x} \bullet \sin{4x} \bullet \sin x = 0\]

\[Первое\ уравнение:\]

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

\[3x = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n.\]

\[x = \frac{1}{3} \bullet \left( \frac{\pi}{2} + \pi n \right) = \frac{\pi}{6} + \frac{\text{πn}}{3}\]

\[Второе\ уравнение:\]

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

\[4x = \arcsin 0 + \pi n = \pi n\]

\[x = \frac{1}{4} \bullet \pi n = \frac{\text{πn}}{4}.\]

\[Третье\ уравнение:\]

\[\sin x = 0\]

\[x = \arcsin 0 + \pi n = \pi n.\]

\[Ответ:\ \ \frac{\pi}{6} + \frac{\text{πn}}{3};\ \ \frac{\text{πn}}{4}\text{.\ }\]

\[2)\sin x \bullet \sin{5x} - \sin^{2}{5x} = 0\]

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

\[\sin{5x} \bullet 2 \bullet \sin\frac{x - 5x}{2} \bullet\]

\[\bullet \cos\frac{x + 5x}{2} = 0\]

\[2 \bullet \sin{5x} \bullet \sin\left( - \frac{4x}{2} \right) \bullet\]

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

\[- 2 \bullet \sin{5x} \bullet \sin{2x} \bullet \cos{3x} = 0\]

\[Первое\ уравнение:\]

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

\[5x = \arcsin 0 + \pi n = \pi n\]

\[x = \frac{1}{5} \bullet \pi n = \frac{\text{πn}}{5}.\]

\[Второе\ уравнение:\]

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

\[2x = \arcsin 0 + \pi n = \pi n\]

\[x = \frac{1}{2} \bullet \pi n = \frac{\text{πn}}{2}.\]

\[Третье\ уравнение:\]

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

\[3x = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n\]

\[x = \frac{1}{3} \bullet \left( \frac{\pi}{2} + \pi n \right) = \frac{\pi}{6} + \frac{\text{πn}}{3}.\]

\[Ответ:\ \ \frac{\text{πn}}{5};\ \ \frac{\text{πn}}{2};\ \ \frac{\pi}{6} + \frac{\text{πn}}{3}\text{.\ }\]