Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1431

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\[1)\ \left\{ \begin{matrix} \sin x \bullet \cos y = - \frac{1}{2} \\ tg\ x \bullet ctg\ y = 1\ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} \sin x \bullet \cos y = - \frac{1}{2} \\ \frac{\sin x}{\cos x} \bullet \frac{\cos y}{\sin y} = 1\ \ \ \ \\ \end{matrix} \right.\ \]

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

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

\[\sin x \bullet \cos y - \cos x \bullet \sin y =\]

\[= - \frac{1}{2} - \left( - \frac{1}{2} \right)\]

\[\sin(x - y) = 0\]

\[x - y = \arcsin 0 + \pi n = \pi n\]

\[x = \pi n + y.\]

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

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

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

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

\[\sin{2y} = \pm 1\]

\[2y = \pm \arcsin 1 + 2\pi m\]

\[2y = \pm \frac{\pi}{2} + 2\pi m\]

\[y = \frac{1}{2} \bullet \left( \pm \frac{\pi}{2} + 2\pi m \right)\]

\[y = \pm \frac{\pi}{4} + \pi m;\]

\[x_{1} = \pi n + \left( - \frac{\pi}{4} + \pi m \right) =\]

\[= - \frac{\pi}{4} + \pi(n + m);\]

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

\[= \frac{\pi}{4} + \pi(n + m);\]

\[Ответ:\ \ \]

\[\left( - \frac{\pi}{4} + \pi(n + m);\ - \frac{\pi}{4} + \pi m \right);\ \ \]

\[\left( \frac{\pi}{4} + \pi(n + m);\ \frac{\pi}{4} + \pi m \right)\text{.\ }\]

\[2)\ \left\{ \begin{matrix} \sin x \bullet \sin y = \frac{1}{4} \\ 3\ tg\ x = ctg\ y\ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} \sin x \bullet \sin y = \frac{1}{4} \\ tg\ x \bullet tg\ y = \frac{1}{3}\text{\ \ } \\ \end{matrix} \right.\ \text{\ \ \ }\]

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

\[1)\ \frac{\frac{1}{4}}{\cos x \bullet \cos y} = \frac{1}{3}\]

\[\cos x \bullet \cos y = \frac{3}{4}\]

\[\cos x \bullet \cos y - \sin x \bullet \sin y = \frac{3}{4} - \frac{1}{4}\]

\[\cos(x + y) = \frac{1}{2}\]

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

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

\[x = \pm \frac{\pi}{3} - y + 2\pi n.\]

\[2)\ \sin x \bullet \sin y = \frac{1}{4}\]

\[\frac{1}{2} \bullet \left( \cos(x - y) - \cos(x + y) \right) = \frac{1}{4}\]

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

\[\cos(x - y) = 1\]

\[x - y = \arccos 1 + 2\pi m = 2\pi m\]

\[x = 2\pi m + y.\]

\[Получим:\]

\[\pm \frac{\pi}{3} - y + 2\pi n = 2\pi m + y\]

\[- 2y = \pm \frac{\pi}{3} + 2\pi m - 2\pi n\]

\[y = \pm \frac{\pi}{6} + \pi(n - m);\]

\[x_{1} = 2\pi m + \left( - \frac{\pi}{6} + \pi n - \pi m \right) =\]

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

\[x_{2} = 2\pi m + \left( \frac{\pi}{6} + \pi n - \pi m \right) =\]

\[= \frac{\pi}{6} + \pi(n + m).\]

\[Ответ:\ \ \]

\[\left( \frac{\pi}{6} + \pi(n + m);\ \frac{\pi}{6} + \pi(n - m) \right);\ \ \]

\[\left( - \frac{\pi}{6} + \pi(n + m);\ - \frac{\pi}{6} + \pi(n - m) \right).\]