ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1218

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

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

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

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

\[1)\ y = \frac{\pi}{2} + \pi n - y - 2\pi n\]

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

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

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

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

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

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

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

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

\[= - \frac{1}{2} + \frac{1}{2}\]

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

\[x + y = \pi n\]

\[2)\ \sin x\cos y - \cos x\sin y = - 1\]

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

\[\sin{(y - x}) = 1\]

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

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

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

\[\left\{ \begin{matrix} x = - \frac{\pi}{4} - \frac{\text{πn}}{2} - \pi k \\ y = \frac{\pi}{4} + \frac{\text{πn}}{2} + \pi k\ \ \ \ \ \\ \end{matrix} \right.\ \]

\[3)\ \left\{ \begin{matrix} \sin x\cos y = \frac{1}{2} \\ \cos x\sin y = \frac{1}{2} \\ \end{matrix} \right.\ \]

\[1)\sin x\cos y + \cos x\text{sin\ }{y = 1}\]

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

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

\[2)\sin x\cos y - \cos x\sin y = 0\]

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

\[x - y = \pi k.\]

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

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

\[\left\{ \begin{matrix} x = \frac{\pi}{4} + \pi n + \frac{\text{πk}}{2} \\ y = \frac{\pi}{4} + \pi n - \frac{\text{πk}}{2} \\ \end{matrix} \right.\ \]

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

\[1)\ \sin x\sin y + \cos x\cos y = \frac{\sqrt{3}}{2}\]

\[\cos(x - y) = \frac{\sqrt{3}}{2}\]

\[x - y = \pm \frac{\pi}{6} + 2\pi n.\]

\[2)\ \sin x\sin y - \cos x\cos y = 0\]

\[\cos(x + y) = 0\]

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

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

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

\[\left\{ \begin{matrix} x = \frac{\pi}{4} \pm \frac{\pi}{12} + \frac{\text{πk}}{2} + \pi n \\ y = \frac{\pi}{4} \pm \frac{\pi}{12} + \frac{\text{πk}}{2} - \pi n \\ \end{matrix} \right.\ \ \]