ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 955

\[1)\ \left\{ \begin{matrix} \sin x + \cos y = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \sin^{2}x + 2\sin x \bullet \cos y = \frac{3}{4} \\ \end{matrix} \right.\ \]

\[\sin x + \cos y = 1\]

\[\cos y = 1 - \sin x.\]

\[\sin^{2}x + 2\sin x \bullet \cos y = \frac{3}{4}\]

\[\sin^{2}x + 2\sin x \bullet \left( 1 - \sin x \right) = \frac{3}{4}\]

\[\sin^{2}x + 2\sin x - 2\sin^{2}x - \frac{3}{4} = 0\]

\[4\sin^{2}x - 8\sin x + 3 = 0\]

\[D = 64 - 48 = 16\]

\[\sin x_{1} = \frac{8 - 4}{2 \bullet 4} = \frac{1}{2};\]

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

\[\sin x = \frac{1}{2};\]

\[\cos y = 1 - \frac{1}{2} = \frac{1}{2};\]

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

\[= ( - 1)^{n}\frac{\pi}{6} + \pi n;\]

\[y = \pm \arccos\frac{1}{2} + 2\pi k =\]

\[= \pm \frac{\pi}{3} + 2\pi k.\]

\[Ответ:\ \ \]

\[\left( ( - 1)^{n}\frac{\pi}{6} + \pi n;\ \pm \frac{\pi}{3} + 2\pi k \right).\]

\[\sin x + \sin y = \frac{1}{2}\]

\[\sin y = \frac{1}{2} - \sin x.\]

\[\cos^{2}x + 2\sin x \bullet \sin y + 4\cos^{2}y = 4\]

\[7\sin^{2}x - 5\sin x = 0\]

\[\sin x \bullet \left( 7\sin x - 5 \right) = 0\]

\[\sin x_{1} = 0;\ \ \ \sin x_{2} = \frac{5}{7};\]

\[\sin y_{1} = \frac{1}{2};\]

\[\sin y_{2} = \frac{1}{2} - \frac{5}{7} = - \frac{3}{14};\]

\[x_{1} = \pi n;\]

\[y_{1} = ( - 1)^{k} \bullet \arcsin\frac{1}{2} + \pi k =\]

\[= ( - 1)^{k}\frac{\pi}{6} + \pi k;\]

\[x_{2} = ( - 1)^{n} \bullet \arcsin\frac{5}{7} + \pi n;\]

\[y_{2} = ( - 1)^{k + 1} \bullet \arcsin\frac{3}{14} + \pi k.\]

\[Ответ:\ \ \]

\[\left( \pi n;\ ( - 1)^{k}\frac{\pi}{6} + \pi k \right)\text{.\ }\]