Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1379

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\[1)\sin{5x} = \sin{3x}\]

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

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

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

\[\sin x = 0\]

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

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

\[4x = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n\]

\[x = \frac{1}{4} \bullet \left( \frac{\pi}{2} + \pi n \right) = \frac{\pi}{8} + \frac{\text{πn}}{4}.\]

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

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

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

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

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

\[4x = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n\]

\[x = \frac{1}{4} \bullet \left( \frac{\pi}{2} + \pi n \right) = \frac{\pi}{8} + \frac{\text{πn}}{4}.\]

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

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

\[x = \frac{1}{2} \bullet \left( \frac{\pi}{2} + \pi n \right) = \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

\[Ответ:\ \ \frac{\pi}{8} + \frac{\text{πn}}{4};\ \ \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

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

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

\[2 \bullet \sin\frac{\frac{\pi}{2} + 7x + 3x}{2} \bullet \cos\frac{\frac{\pi}{2} + 7x - 3x}{2} = 0\]

\[\sin\left( \frac{\pi}{4} + 5x \right) \bullet \cos\left( \frac{\pi}{4} + 2x \right) = 0\]

\[\sin\left( \frac{\pi}{4} + 5x \right) = 0\]

\[\frac{\pi}{4} + 5x = \arcsin 0 + \pi n = \pi n\]

\[5x = - \frac{\pi}{4} + \pi n\]

\[x = \frac{1}{5} \bullet \left( - \frac{\pi}{4} + \pi n \right)\]

\[x = - \frac{\pi}{20} + \frac{\text{πn}}{5}.\]

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

\[\frac{\pi}{4} + 2x = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n\]

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

\[x = \frac{1}{2} \bullet \left( \frac{\pi}{4} + \pi n \right) = \frac{\pi}{8} + \frac{\text{πn}}{2}.\]

\[Ответ:\ \ - \frac{\pi}{20} + \frac{\text{πn}}{5};\ \ \frac{\pi}{8} + \frac{\text{πn}}{2}.\]

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

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

\[\sin x - \sin\left( \frac{\pi}{2} - 5x \right) = 0\]

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

\[\cos\left( \frac{\pi}{4} - 2x \right) \bullet \sin\left( 3x - \frac{\pi}{4} \right) = 0\]

\[\cos\left( 2x - \frac{\pi}{4} \right) = 0\]

\[2x - \frac{\pi}{4} = \arccos 0 + \pi n\]

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

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

\[2x = \frac{3\pi}{4} + \pi n\]

\[x = \frac{1}{2} \bullet \left( \frac{3\pi}{4} + \pi n \right)\]

\[x = \frac{3\pi}{8} + \frac{\text{πn}}{2}.\]

\[\sin\left( 3x - \frac{\pi}{4} \right) = 0\]

\[3x - \frac{\pi}{4} = \arcsin 0 + \pi n = \pi n\]

\[3x = \frac{\pi}{4} + \pi n\]

\[x = \frac{1}{3} \bullet \left( \frac{\pi}{4} + \pi n \right)\]

\[x = \frac{\pi}{12} + \frac{\text{πn}}{3}.\]

\[Ответ:\ \ \frac{3\pi}{8} + \frac{\text{πn}}{2};\ \ \frac{\pi}{12} + \frac{\text{πn}}{3}.\]