Страница 427 номер 1572, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

\(\begin{cases} 6\sin x\cos y + 2\cos x\sin y = - 3 \\ 5\sin x\cos y - 3\cos x\sin y = 1 \end{cases}\)
\(\sin x\cos y = z, \cos x\sin y = w\)
\(\begin{cases} 6z + 2w = - 3 \\ 5z - 3w = 1 \end{cases},\)
\(\begin{cases} z = \frac{-3 - 2w}{6} \\ 5\left( \frac{-3 - 2w}{6} \right) - 3w = 1 \end{cases},\)
\(- 15 - 10w - 18w - 6 = 0,\)
\(- 28w = 21,\)
\(w = - \dfrac{3}{4},\)
\(z = \dfrac{-3 + 2 \cdot \frac{3}{4}}{6} = - \dfrac{1}{4};\)
\(\begin{cases} \sin x\cos y = - \frac{1}{4} \\ \cos x\sin y = - \frac{3}{4} \end{cases},\)
\(\begin{cases} 4\sin x\cos y = - 1 \\ 4\cos x\sin y = - 3 \end{cases},\)
\(4\sin(x + y) = - 4,\)
\(\sin(x + y) = - 1,\)
\(x + y = - \dfrac{\pi}{2} + 2\pi n,\)
\(x = - y - \dfrac{\pi}{2} + 2\pi n, n \in Z;\)
\(4\sin(x - y) = 2,\)
\(\sin(x - y) = \dfrac{1}{2},\)
\(x - y = (-1)^{m} \cdot \dfrac{\pi}{6} + \pi m, m \in Z;\)
\(- 2y - \dfrac{\pi}{2} + 2\pi n = (-1)^{m} \cdot \dfrac{\pi}{6} + \pi m,\)
\(- y = (-1)^{m} \cdot \dfrac{\pi}{12} + \dfrac{\pi m}{2} + \dfrac{\pi}{4} - \pi n,\)
\(y = (-1)^{m + 1} \cdot \dfrac{\pi}{12} - \dfrac{\pi}{4} - \dfrac{\pi m}{2} + \pi n,\)
\(x = (-1)^{m} \cdot \dfrac{\pi}{12} + \dfrac{\pi m}{2} + \dfrac{\pi}{4} - \pi n - \dfrac{\pi}{2} + 2\pi n =\)
\(= (-1)^{m} \cdot \dfrac{\pi}{12} - \dfrac{\pi}{4} + \dfrac{\pi m}{2} + \pi n, n,m \in Z.\)