Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 724

\[\boxed{\mathbf{724}\mathbf{.}}\]

\[1)\sin x = \frac{\sqrt{3}}{2}\]

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

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

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

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

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

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

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

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

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

\[= ( - 1)^{n} \bullet \frac{\pi}{4} + \pi n;\]

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

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

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

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

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

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

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

\[= ( - 1)^{n + 1} \bullet \frac{\pi}{4} + \pi n;\]

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

\[x_{1} = \frac{\pi}{4} + \pi = \frac{5\pi}{4};\]

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

\[4)\sin x = - \frac{\sqrt{3}}{2};\]

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

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

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

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

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