Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1383

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\[1)\sin x + \sin{2x} + \sin{3x} + \sin{4x} = 0\]

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

\[\sin\frac{5x}{2} \bullet \left( \cos\frac{3x}{2} + \cos\frac{x}{2} \right) = 0\]

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

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

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

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

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

\[\cos x = 0\]

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

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

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

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

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

\[2)\cos x + \cos{2x} + \cos{3x} + \cos{4x} = 0\]

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

\[\cos\frac{5x}{2} \bullet \left( \cos\frac{3x}{2} + \cos\frac{x}{2} \right) = 0\]

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

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

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

\[\frac{5x}{2} = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n\]

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

\[\cos x = 0\]

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

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

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

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

\[Ответ:\ \ \]

\[\frac{\pi}{5} + \frac{2\pi n}{5};\ \ \frac{\pi}{2} + \pi n;\ \ \pi + 2\pi n.\]