ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 962

\[\left\{ \begin{matrix} 6\sin x\cos y + 2\cos x\sin y = - 3 \\ 5\sin x\cos y - 3\cos x\sin y = 1\ \ \ \\ \end{matrix} \right.\ \]

\[1)\ u = \sin x\cos y;\ v = \cos x\sin y:\]

\[6u + 2v = - 3\]

\[2v = - 6u - 3\]

\[v = \frac{- 6u - 3}{2}.\]

\[5u - 3v = 1\]

\[3v = 5u - 1\ \]

\[v = \frac{5u - 1}{3}.\]

\[\frac{- 6u - 3}{2} = \frac{5u - 1}{3}\]

\[3( - 6u - 3) = 2(5u - 1)\]

\[- 18u - 9 = 10u - 2\]

\[28u = - 7\]

\[u = - \frac{1}{4};\]

\[v = \frac{\frac{3}{2} - 3}{2} = \frac{3 - 6}{4} = - \frac{3}{4}.\]

\[2)\ v + u = - \frac{1}{4} - \frac{3}{4} = - 1\]

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

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

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

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

\[3)\ v - u = - \frac{1}{4} + \frac{3}{4} = \frac{1}{2}\]

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

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

\[x - y = ( - 1)^{k}\frac{\pi}{6} + \pi k\]

\[x + x + \frac{\pi}{2} - 2\pi n = ( - 1)^{k}\frac{\pi}{6} + \pi k\]

\[2x = - \frac{\pi}{2} + ( - 1)^{k}\frac{\pi}{6} + \pi k + 2\pi n\]

\[x = - \frac{\pi}{4} + ( - 1)^{k}\frac{\pi}{12} + \frac{\text{πk}}{2} + \pi n;\]

\[y = - \frac{\pi}{4} + ( - 1)^{k + 1}\frac{\pi}{12} - \frac{\text{πk}}{2} + \pi n.\]

\[Ответ:\ \ \]