ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1362

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\[\mathbf{На\ }отрезке\ \lbrack - \pi;\ 3\pi\rbrack.\]

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

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

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

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

\[y = \cos x\ и\ y = - \frac{1}{2}:\]

\[x_{1} = - \frac{2\pi}{3};\]

\[x_{2} = \frac{2\pi}{3};\]

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

\[x_{4} = \frac{2\pi}{3} + 2\pi = \frac{8\pi}{3}.\]

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

\[x = ( - 1)^{n + 1} \bullet \arcsin\frac{\sqrt{3}}{2} + \pi n =\]

\[= ( - 1)^{n + 1} \bullet \frac{\pi}{3} + \pi n;\]

\[y = \sin x\ и\ y = - \frac{\sqrt{3}}{2}:\]

\[x_{1} = ( - 1)^{- 1 + 1} \bullet \frac{\pi}{3} - \pi =\]

\[= \frac{\pi}{3} - \pi = - \frac{2\pi}{3};\]

\[x_{2} = ( - 1)^{0 + 1} \bullet \frac{\pi}{3} = - \frac{\pi}{3};\]

\[x_{3} = ( - 1)^{1 + 1} \bullet \frac{\pi}{3} + \pi =\]

\[= \frac{\pi}{3} + \pi = \frac{4\pi}{3};\]

\[x_{4} = ( - 1)^{2 + 1} \bullet \frac{\pi}{3} + 2\pi =\]

\[= - \frac{\pi}{3} + 2\pi = \frac{5\pi}{3}.\]