ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1219

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

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

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

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

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

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

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

\[2\sin y = 1\]

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

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

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

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

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

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

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

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

\[2)\ \left\{ \begin{matrix} \cos x + \cos y = \frac{1}{2}\text{\ \ \ } \\ \sin^{2}x + \sin^{2}y = \frac{7}{4} \\ \end{matrix} \right.\ \]

\[a = \cos x;\ \ b = \cos y;\ \ \]

\[\ a + b = \frac{1}{2}:\]

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

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

\[- \left( \cos^{2}x + \cos^{2}y \right) = \frac{7}{4}\]

\[a^{2} + b^{2} = \frac{1}{4}\]

\[\left\{ \begin{matrix} a + b = \frac{1}{2}\text{\ \ \ \ } \\ a^{2} + b^{2} = \frac{1}{4} \\ \end{matrix} \right.\ \]

\[\frac{1}{4} = a^{2} + b^{2} = (a + b)^{2} - 2ab =\]

\[= \frac{1}{4} - 2ab\]

\[2ab = 0\]

\[a = 0\ или\ b = 0.\]

\[\left\{ \begin{matrix} \cos x = \frac{1}{2} \\ \cos y = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }или\ \ \ \ \ \ \left\{ \begin{matrix} \cos x = 0 \\ \cos y = \frac{1}{2} \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = \pm \frac{\pi}{3} + 2\pi k \\ y = \frac{\pi}{2} + 2\pi n\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = \frac{\pi}{2} + 2\pi k\ \ \ \\ y = \pm \frac{\pi}{3} + 2\pi n \\ \end{matrix} \right.\ \]

\[3)\ \left\{ \begin{matrix} \cos(x - y) = 2\cos(x + y) \\ \cos x\cos y = \frac{3}{4}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[\cos(x - y) - 2\cos(x + y) =\]

\[= - \cos x\cos y + 3\sin x\sin y =\]

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

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

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

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

\[- \sin x\sin y = \frac{3}{4} - \frac{1}{4} = \frac{1}{2}\]

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

\[2)\cos(x - y) = \cos x\cos y +\]

\[+ \sin x\sin y = \frac{3}{4} + \frac{1}{4} = 1.\]

\[x - y = \pi + 2\pi n.\]

\[\left\{ \begin{matrix} x + y = \pm \frac{\pi}{3} + 2\pi k \\ x - y = \pi + 2\pi n\ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 2x = \pi \pm \frac{\pi}{3} + 2\pi n + 2\pi k\ \ \ \ \\ 2y = - \pi \pm \frac{\pi}{3} + 2\pi n + 2\pi k \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = \frac{\pi}{2} \pm \frac{\pi}{6} + \pi n + \pi k\ \ \ \ \ \\ y = - \frac{\pi}{2} + \frac{\pi}{6} + \pi n + \pi k \\ \end{matrix} \right.\ \]

\[4)\ \left\{ \begin{matrix} \cos\frac{x + y}{2}\cos\frac{x - y}{2} = \frac{1}{2} \\ \cos x\cos y = \frac{1}{4}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[\cos\frac{x + y}{2}\cos\frac{x - y}{2} =\]

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

\[= \frac{1 + \cos x}{2} \cdot \frac{1 + \cos y}{2} -\]

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

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

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

\[a = \cos x;\ \ b = \cos y;\ \ \]

\[\ a + b = 1.\]

\[a \cdot b = \frac{1}{4}\]

\[a \cdot (1 - a) = \frac{1}{4}\]

\[a^{2} - a + \frac{1}{4} = 0\]

\[\left( a - \frac{1}{2} \right)^{2} = 0\]

\[a = \frac{1}{2};\]

\[b = 1 - \frac{1}{2} = \frac{1}{2}.\]

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

\[\left\{ \begin{matrix} x = \pm \frac{\pi}{3} + 2\pi n \\ y = \pm \frac{\pi}{3} + 2\pi k \\ \end{matrix} \right.\ \]