Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 645

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\[1)\ \left\{ \begin{matrix} \cos(x + y) = 0 \\ \cos(x - y) = 1 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x + y = \arccos 0 + \pi n\ \ \\ x - y = \arccos 1 + 2\pi n \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} x + y = \frac{\pi}{2} + \pi n \\ x - y = 2\pi n\ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} x = \frac{\pi}{2} + \pi n - y \\ y = x - 2\pi n\ \ \ \ \ \\ \end{matrix} \right.\ \]

\[y = \frac{\pi}{2} + \pi n - y - 2\pi n\]

\[2y = \frac{\pi}{2} - \pi n\]

\[y = \frac{1}{2} \bullet \left( \frac{\pi}{2} - \pi n \right) = \frac{\pi}{4} - \frac{\text{πn}}{2}.\]

\[x = \frac{\pi}{2} + \pi n - \frac{\pi}{4} + \frac{\text{πn}}{2}\]

\[x = \frac{\pi}{4} + \frac{3\pi n}{2}.\]

\[Ответ:\ \ x = \frac{\pi}{4} + \frac{3\pi n}{2};\ \ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }y = \frac{\pi}{4} - \frac{\text{πn}}{2}.\]

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

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

\[1 + \sin y - \sin y =\]

\[= \left( 1 + \sin y \right)^{2} + \cos^{2}y\]

\[1 = 1 + 2\sin y + \sin^{2}y + \cos^{2}y\]

\[1 + 2\sin y + 1 - 1 = 0\]

\[2\sin y + 1 = 0\]

\[2\sin y = 1\]

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

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

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

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

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

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

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

\[\text{\ \ \ \ \ \ \ \ \ \ }y = ( - 1)^{n + 1} \bullet \frac{\pi}{6} + \pi n.\ \]