Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 712

\[\boxed{\mathbf{712}\mathbf{.}}\]

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

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

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

\[\lbrack 0;\ 3\pi\rbrack:\]

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

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

\[x_{3} = \frac{\pi}{3} + 2\pi = \frac{7\pi}{3}.\]

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

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

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

\[\lbrack 0;\ 3\pi\rbrack:\]

\[x_{1} = \frac{\pi}{4};\]

\[x_{2} = - \frac{\pi}{4} + 2\pi = \frac{7\pi}{4};\]

\[x_{3} = \frac{\pi}{4} + 2\pi = \frac{9\pi}{4}.\ \]

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

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

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

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

\[\lbrack 0;\ 3\pi\rbrack:\]

\[x_{1} = \frac{3\pi}{4};\]

\[x_{2} = - \frac{3\pi}{4} + 2\pi = \frac{5\pi}{4};\]

\[x_{3} = \frac{3\pi}{4} + 2\pi = \frac{11\pi}{4}.\]

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

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

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

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

\[\lbrack 0;\ 3\pi\rbrack:\]

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

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

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