Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 715

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

\[2x = \pm \arccos\frac{1}{2} + 2\pi n\]

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

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

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

\[\left\lbrack - \frac{\pi}{2};\ \frac{3\pi}{2} \right\rbrack:\]

\[x_{1} = - \frac{\pi}{6};\]

\[x_{2} = \frac{\pi}{6};\]

\[x_{3} = - \frac{\pi}{6} + \pi = \frac{5\pi}{6};\]

\[x_{4} = \frac{\pi}{6} + \pi = \frac{7\pi}{6}.\]

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

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

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

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

\[x = \pm \frac{\pi}{18} + \frac{2\pi n}{3};\]

\[\left\lbrack - \frac{\pi}{2};\ \frac{3\pi}{2} \right\rbrack:\]

\[x_{1} = - \frac{\pi}{18};\]

\[x_{2} = \frac{\pi}{18};\]

\[x_{3} = - \frac{\pi}{18} + \frac{2\pi}{3} = \frac{11\pi}{18};\]

\[x_{4} = \frac{\pi}{18} + \frac{2\pi}{3} = \frac{13\pi}{18};\]

\[x_{5} = - \frac{\pi}{18} + \frac{4\pi}{3} = \frac{23\pi}{18};\]

\[x_{6} = \frac{\pi}{18} + \frac{4\pi}{3} = \frac{25\pi}{18}.\]