Страница 415 номер 1430, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\begin{cases} \sin x + \cos y = 1 \\ \sin^{2}x + 2\sin x\cos y = \frac{3}{4} \end{cases}\)
\(\begin{cases} \cos y = 1 - \sin x \\ \sin^{2}x + 2\sin x - 2\sin^{2}x = \frac{3}{4} \end{cases},\)
\(\begin{cases} \cos y = 1 - \sin x \\ \sin^{2}x - 2\sin x + \frac{3}{4} = 0 \end{cases},\)
\(4\sin^{2}x - 8\sin x + 3 = 0,\)
\(\sin x = z, 4z^{2} - 8z + 3 = 0\)
\(z_{1{,}2} = \dfrac{8 \pm \sqrt{64 - 48}}{8},\)
\(z_{1} = \dfrac{3}{2}, z_{2} = \dfrac{1}{2}\)
\(\sin x = \frac{3}{2}\) – решений нет;
\(\begin{cases} \cos y = 1 - \frac{1}{2} \\ \sin x = \frac{1}{2} \end{cases},\)
\(\begin{cases} y = \pm \frac{\pi}{3} + 2\pi m \\ x = (-1)^{n} \cdot \frac{\pi}{6} + \pi n \end{cases}, n,m \in Z;\)

2) \(\begin{cases} \sin x + \sin y = \frac{1}{2} \\ \cos^{2}x + 2\sin x\sin y + 4\cos^{2}y = 4 \end{cases}\)
\(\begin{cases} \sin x = \frac{1}{2} - \sin y \\ 1 - \sin^{2}x + 2\left( \frac{1}{2} - \sin y \right)\sin y + 4\cos^{2}y = 4 \end{cases},\)
\(1 - \dfrac{1}{4} + \sin y - \sin^{2}y + \sin y - 2\sin^{2}y + 4\cos^{2}y - 4 = 0,\)
\(\dfrac{3}{4} + 2\sin y - 7\sin^{2}y = 0,\)
\(\sin y = z, 28z^{2} - 8z - 3 = 0\)
\(z_{1{,}2} = \dfrac{8 \pm \sqrt{64 + 336}}{56},\)
\(z_{1} = \dfrac{1}{2}, z_{2} = - \dfrac{3}{14}\)
\(\begin{cases} \sin y = \frac{1}{2} \\ \sin x = \frac{1}{2} - \frac{1}{2} \end{cases}\) или \(\begin{cases} \sin y = - \frac{3}{14} \\ \sin x = \frac{1}{2} + \frac{3}{14} \end{cases}\)
\(\begin{cases} y = (-1)^{n} \cdot \frac{\pi}{6} + \pi n \\ x = \pi m \end{cases}, n,m \in Z;\)
\(\begin{cases} y = (-1)^{n + 1} \cdot \arcsin\frac{3}{14} + \pi n \\ x = (-1)^{m}\arcsin\frac{5}{7} + \pi m \end{cases}, n,m \in Z.\)