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МатематикаГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 956
Задание 956
\[1)\ \left\{ \begin{matrix} \sin x \bullet \cos y = - \frac{1}{2} \\ tg\ x \bullet ctg\ y = 1\ \ \ \ \\ \end{matrix} \right.\ \]
\[tg\ x \bullet ctg\ y = 1\]
\[\frac{\sin x}{\cos x} \bullet \frac{\cos y}{\sin y} = 1\]
\[- \frac{1}{2}\ :\left( \cos x \bullet \sin y \right) = 1\]
\[- 2\cos x \bullet \sin y = 1\]
\[- \cos x \bullet \sin y = \frac{1}{2}\]
\[\sin x \bullet \cos y - \cos x \bullet \sin y = - \frac{1}{2} + \frac{1}{2}\]
\[\sin(x - y) = 0\]
\[x - y = \pi n\]
\[x = \pi n + y.\]
\[\sin x \bullet \cos y = - \frac{1}{2}\]
\[\sin x \bullet \cos y = - \frac{1}{2}\]
\[\frac{1}{2}\left( \sin(x + y) + \sin(x - y) \right) = - \frac{1}{2}\]
\[\sin(x + y) = - 1\]
\[x + y = - \frac{\pi}{2} + 2\pi k\]
\[x = - \frac{\pi}{2} - y + 2\pi k.\]
\[Получим:\]
\[\pi n + y = - \frac{\pi}{2} - y + 2\pi k\]
\[2y = - \frac{\pi}{2} + \pi(2k - n)\]
\[y = - \frac{\pi}{4} + \pi\left( k - \frac{n}{2} \right)\]
\[x = \pi n - \frac{\pi}{4} + \pi\left( k - \frac{n}{2} \right) =\]
\[= - \frac{\pi}{4} + \pi\left( k + \frac{n}{2} \right).\]
\[Ответ:\ \ \]
\[\left( - \frac{\pi}{4} + \pi\left( k + \frac{n}{2} \right);\ - \frac{\pi}{4} + \pi\left( k - \frac{n}{2} \right) \right)\text{.\ }\]
\[2)\ \left\{ \begin{matrix} \sin x \bullet \sin y = \frac{1}{4} \\ 3\ tg\ x = ctg\ y\ \ \\ \end{matrix} \right.\ \]
\[3\ tg\ x = ctg\ y\]
\[3\frac{\sin x}{\cos x} = \frac{\cos y}{\sin y}\]
\[\frac{\sin x \bullet \sin y}{\cos x \bullet \cos y} = \frac{1}{3}\]
\[\frac{1}{4}\ :\left( \cos x \bullet \cos y \right) = \frac{1}{3}\]
\[4\cos x \bullet \cos y = 3\]
\[\cos x \bullet \cos y = \frac{3}{4}\]
\[\cos x \bullet \cos y - \sin x \bullet \sin y = \frac{3}{4} - \frac{1}{4}\]
\[\cos(x + y) = \frac{1}{2}\]
\[x + y = \pm \arccos\frac{1}{2} + 2\pi n = \pm \frac{\pi}{3} + 2\pi n\]
\[x = \pm \frac{\pi}{3} - y + 2\pi n.\]
\[\sin x \bullet \sin y = \frac{1}{4}\]
\[\sin x \bullet \sin y = \frac{1}{4}\]
\[\frac{1}{2} \bullet \left( \cos(x - y) - \cos(x + y) \right) = \frac{1}{4}\]
\[\cos(x - y) - \frac{1}{2} = \frac{1}{2}\]
\[\cos(x - y) = 1\]
\[x - y = 2\pi k\]
\[x = 2\pi k + y.\]
\[Получим:\]
\[\pm \frac{\pi}{3} - y + 2\pi n = 2\pi k + y\]
\[- 2y = \pm \frac{\pi}{3} + 2\pi k - 2\pi n\]
\[y = \pm \frac{\pi}{6} + \pi(n - k)\]
\[x_{1} = 2\pi k - \frac{\pi}{6} + \pi n - \pi k = - \frac{\pi}{6} + \pi(n + k);\]
\[x_{2} = 2\pi k + \frac{\pi}{6} + \pi n - \pi k = \frac{\pi}{6} + \pi(n + k).\]
\[Ответ:\ \ \]
\[\left( \frac{\pi}{6} + \pi(n + k);\ \frac{\pi}{6} + \pi(n - k) \right);\ \]
\[\left( - \frac{\pi}{6} + \pi(n + k);\ - \frac{\pi}{6} + \pi(n - k) \right).\]
