Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 635

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\[1)\cos x \bullet \cos{2x} = \sin x \bullet \sin{2x}\]

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

\[\cos(x + 2x) = 0\]

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

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

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

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

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

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

\[\sin(2x - x) = 0\]

\[\sin x = 0\]

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

\[Ответ:\ \ \pi n.\]

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

\[\sin(x + 2x) - \sin{2x} \bullet \cos x = 0\]

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

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

\[1)\ \sin x = 0\]

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

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

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

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

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

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

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

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

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

\[1)\ \sin{5x} = 0\]

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

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

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

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

\[Ответ:\ \ \frac{\text{πn}}{5}.\]