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МатематикаРешебник по алгебре и начала математического анализа 10 класс Алимов Задание 1430
Задание 1430
\[\boxed{\mathbf{1430}\mathbf{.}}\]
\[1)\ \left\{ \begin{matrix} \sin x + \cos y = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \sin^{2}x + 2\sin x \bullet \cos y = \frac{3}{4} \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} \cos y = 1 - \sin x\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \sin^{2}x + 2\sin x \bullet \cos y - \frac{3}{4} = 0 \\ \end{matrix} \right.\ \]
\[\sin^{2}x + 2\sin x \bullet \left( 1 - \sin x \right) - \frac{3}{4} = 0\]
\[\sin^{2}x + 2\sin x - 2\sin^{2}x - \frac{3}{4} = 0\ \ \ \ \ | \bullet ( - 4)\]
\[4\sin^{2}x - 8\sin x + 3 = 0\]
\[a = \sin x:\]
\[4a^{2} - 8a + 3 = 0\]
\[D = 64 - 48 = 16\]
\[a_{1} = \frac{8 - 4}{2 \bullet 4} = \frac{4}{8} = \frac{1}{2};\]
\[a_{2} = \frac{8 + 4}{2 \bullet 4} = \frac{12}{8} = \frac{3}{2}.\]
\[1)\ \sin x = \frac{1}{2}\]
\[x = ( - 1)^{n} \bullet \arcsin\frac{1}{2} + \pi n =\]
\[= ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n;\]
\[\cos y = 1 - \frac{1}{2} = \frac{1}{2}\]
\[y = \pm \arccos\frac{1}{2} + 2\pi n =\]
\[= \pm \frac{\pi}{3} + 2\pi n.\]
\[2)\ \sin x = \frac{3}{2}\]
\[корней\ нет.\]
\[Ответ:\ \ \]
\[\left( ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n;\ \ \pm \frac{\pi}{3} + 2\pi n \right).\]
\[2)\ \left\{ \begin{matrix} \sin x + \sin y = \frac{1}{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \cos^{2}x + 2\sin x \bullet \sin y + 4\cos^{2}y = 4 \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\sin x = \frac{1}{2} - \sin y;\]
\[\frac{3}{4} + 2\sin y - 7\sin^{2}y = 0\ \ \ \ \ | \bullet ( - 4)\]
\[28\sin^{2}y - 8\sin y - 3 = 0\]
\[a = \sin y:\]
\[28a^{2} - 8a - 3 = 0\]
\[D = 64 + 336 = 400\]
\[a_{1} = \frac{8 - 20}{2 \bullet 28} = - \frac{12}{56} = - \frac{3}{14};\]
\[a_{2} = \frac{8 + 20}{2 \bullet 28} = \frac{28}{56} = \frac{1}{2}.\]
\[1)\ \sin y = - \frac{3}{14}\]
\[y = ( - 1)^{n + 1} \bullet \arcsin\frac{3}{14} + \pi n;\]
\[\sin x = \frac{1}{2} + \frac{3}{14} = \frac{5}{7}\]
\[x = ( - 1)^{n} \bullet \arcsin\frac{5}{7} + \pi n.\]
\[2)\ \sin y = \frac{1}{2}\]
\[y = ( - 1)^{n} \bullet \arcsin\frac{1}{2} + \pi n =\]
\[= ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n;\]
\[\sin x = \frac{1}{2} - \frac{1}{2} = 0\]
\[x = \arcsin 0 + \pi n = \pi n.\]
\[Ответ:\ \ \]
\[\left( \pi n;\ \ ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n \right)\text{.\ }\]
