Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 626

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

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

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

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

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

\[1)\sin{2x} = 0\]

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

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

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

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

\[Ответ:\ \ \frac{\text{πn}}{2}.\]

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

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

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

\[\sin\frac{4x}{2} \bullet \cos\frac{6x}{2} = 0\]

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

\[1)\ \sin{2x} = 0\]

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

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

\[2)\ \cos{3x} = 0\]

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

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

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

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

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

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

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

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

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

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

\[1)\ \sin\left( \frac{5x}{2} - \frac{\pi}{4} \right) = 0\]

\[\frac{5x}{2} - \frac{\pi}{4} = \arcsin 0 + \pi n = \pi n\]

\[\frac{5x}{2} = \pi n + \frac{\pi}{4}\]

\[x = \frac{2}{5} \bullet \left( \frac{\pi}{4} + \pi n \right) = \frac{\pi}{10} + \frac{2\pi n}{5}.\]

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

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

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

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

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

\[Ответ:\ \ \frac{\pi}{10} + \frac{2\pi n}{5};\ \ - \frac{\pi}{2} + 2\pi n.\]

\[4)\sin x + \cos{3x} = 0\]

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

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

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

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

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

\[1)\ \cos\left( 2x - \frac{\pi}{4} \right) = 0\]

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

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

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

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

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

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

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

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

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

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

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

\[Ответ:\ \ \frac{3\pi}{8} + \frac{\text{πn}}{2};\ \ \frac{\pi}{4} + \pi n.\]