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МатематикаРешебник по алгебре и начала математического анализа 10 класс Колягин Задание 1261
Задание 1261
\[\boxed{\mathbf{1261}\mathbf{.}}\]
\[1)\ \left\{ \begin{matrix} \sin y \bullet \cos y = \frac{1}{2}\text{\ \ \ \ \ \ \ } \\ \sin{2x} + \sin{2y} = 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} \frac{1}{2}\sin{2y} = \frac{1}{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \sin{2x} + \sin{2y} = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} \sin{2y} = 1\ \ \ \ \ \ \ \ \\ \sin{2x} + 1 = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\left\{ \begin{matrix} \sin{2y = 1}\text{\ \ \ } \\ \sin{2x} = - 1 \\ \end{matrix} \right.\ \]
\[1)\ \sin{2x} = - 1\]
\[2x = - \arcsin 1 + 2\pi n =\]
\[= - \frac{\pi}{2} + 2\pi n\]
\[x = \frac{1}{2} \bullet \left( - \frac{\pi}{2} + 2\pi n \right) = - \frac{\pi}{4} + \pi n.\]
\[2)\ \sin{2y} = 1\]
\[2y = \arcsin 1 + 2\pi n = \frac{\pi}{2} + 2\pi n\]
\[y = \frac{1}{2} \bullet \left( \frac{\pi}{2} + 2\pi n \right) = \frac{\pi}{4} + \pi n\]
\[Ответ:\ \ x = - \frac{\pi}{4} + \pi n;\ \]
\[\ y = \frac{\pi}{4} + \pi n.\]
\[2)\ \left\{ \begin{matrix} \sin x + \sin y = 1\ \ \ \ \\ \cos x - \cos y = \sqrt{3} \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 2 \bullet \sin\frac{x + y}{2} \bullet \cos\frac{x - y}{2} = 1\ \ \ \ \ \\ - 2 \bullet \sin\frac{x + y}{2} \bullet \sin\frac{x - y}{2} = \sqrt{3} \\ \end{matrix} \right.\ \]
\[Разделим\ почленно\ второе\ \]
\[уравнение\ на\ первое:\]
\[- tg\frac{x - y}{2} = \sqrt{3}\]
\[\text{tg}\frac{x - y}{2} = - \sqrt{3}\]
\[\frac{x - y}{2} = - arctg\ \sqrt{3} + \pi n =\]
\[= - \frac{\pi}{3} + \pi n\]
\[x - y = 2 \bullet \left( - \frac{\pi}{3} + \pi n \right) =\]
\[= - \frac{2\pi}{3} + 2\pi n\]
\[x = y - \frac{2\pi}{3} + 2\pi n.\]
\[1)\ \sin\left( y - \frac{2\pi}{3} + 2\pi n \right) +\]
\[+ \sin y = 1\]
\[\sin\left( y - \frac{2\pi}{3} \right) + \sin y = 1\]
\[\sin y \bullet \cos\frac{2\pi}{3} - \sin\frac{2\pi}{3} \bullet \cos y +\]
\[+ \sin y = 1\]
\[\sin y \bullet \cos\left( \pi - \frac{\pi}{3} \right) -\]
\[- \sin\left( \pi - \frac{\pi}{3} \right) \bullet \cos y + \sin y = 1\]
\[- \sin y \bullet \cos\frac{\pi}{3} - \sin\frac{\pi}{3} \bullet \cos y +\]
\[+ \sin y = 1\]
\[- \frac{1}{2}\sin y - \frac{\sqrt{3}}{2}\cos y + \sin y = 1\]
\[\frac{1}{2}\sin y - \frac{\sqrt{3}}{2}\cos y = 1\]
\[\cos\frac{\pi}{3} \bullet \sin y - \sin\frac{\pi}{3} \bullet \cos y = 1\]
\[\sin\left( y - \frac{\pi}{3} \right) = 1\]
\[y - \frac{\pi}{3} = \arcsin 1 + 2\pi k =\]
\[= \frac{\pi}{2} + 2\pi k\]
\[y = \frac{\pi}{2} + \frac{\pi}{3} + 2\pi k = \frac{3\pi}{6} + \frac{2\pi}{6} +\]
\[+ 2\pi k = \frac{5\pi}{6} + 2\pi k.\]
\[2)\ x = y - \frac{2\pi}{3} + 2\pi n = \frac{5\pi}{6} -\]
\[- \frac{4\pi}{6} + 2\pi n + 2\pi k = \frac{\pi}{6} +\]
\[+ 2\pi(n + k).\]
\[Ответ:\ \ x = \frac{\pi}{6} + 2\pi(n + k);\]
\[\ \ y = \frac{5\pi}{6} + 2\pi k.\]
