Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 573

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\[1)\cos{4x} = 1\]

\[4x = \arccos 1 + 2\pi n =\]

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

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

\[x = \frac{\text{πn}}{2}\]

\[Ответ:\ \ \frac{\pi n}{2}.\]

\[2)\cos{2x} = - 1\]

\[2x = \left( \pi - \arccos 1 \right) + 2\pi n =\]

\[= \pi + 2\pi n\]

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

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

\[Ответ:\ \ \frac{\pi}{2} + \pi n.\]

\[3)\ \sqrt{2}\cos\frac{x}{4} = - 1\]

\[\cos\frac{x}{4} = - \frac{1}{\sqrt{2}}\]

\[\frac{x}{4} = \pm \left( \pi - \arccos\frac{1}{\sqrt{2}} \right) + 2\pi n =\]

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

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

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

\[x = \pm 3\pi + 8\pi n\]

\[Ответ:\ \pm 3\pi + 8\pi n.\]

\[4)\ 2\cos\frac{x}{3} = \sqrt{3}\]

\[\cos\frac{x}{3} = \frac{\sqrt{3}}{2}\]

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

\[= \pm \frac{\pi}{6} + 2\pi n\]

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

\[x = \pm \frac{\pi}{2} + 6\pi n\]

\[Ответ:\ \pm \frac{\pi}{2} + 6\pi n.\]

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

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

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

\[x = \frac{3\pi - 2\pi}{6} + \pi n\]

\[x = \frac{\pi}{6} + \pi n\]

\[Ответ:\ \ \frac{\pi}{6} + \pi n.\]

\[6)\cos\left( 2x - \frac{\pi}{4} \right) = 0\]

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

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

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

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

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

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

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