Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1264

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\[1)\ \left\{ \begin{matrix} \text{ct}g^{2}x + 32\sin^{2}y = 35 \\ \frac{1}{\sin^{2}{2x}} - 4\cos y = 5\ \ \ \\ \end{matrix} \right.\ \]

\[\text{ct}g^{2}x + 32\sin^{2}y - \frac{1}{\sin^{2}{2x}} +\]

\[+ 4\cos y = 30\]

\[32{\sin^{2}\ }y + 4\cos y - 31 = 0\]

\[- 32\cos^{2}y + 4\cos y + 1 = 0\]

\[Пусть\cos y = t:\]

\[32t^{2} - 4t - 1 = 0\]

\[D_{1} = 4 + 32 = 36\]

\[t_{1} = \frac{2 + 6}{32} = \frac{1}{4};\ \ \ \ \]

\[\ t_{2} = \frac{2 - 6}{32} = - \frac{1}{8}.\]

\[\cos y = \frac{1}{4}\]

\[y = \pm \arccos\frac{1}{4} + \pi n.\]

\[\cos y = - \frac{1}{8}\]

\[y = \pm \left( \pi - \arccos\frac{1}{8} \right) + \pi n.\]

\[\text{ct}g^{2}2x = 35 - 32\sin^{2}y\]

\[\text{ct}g^{2}2x = 35 -\]

\[- 32\sin^{2}\left( \pm \arccos\frac{1}{4} + \pi n \right) = 5\]

\[\text{ct}g^{2}2x = 35 -\]

\[- 32\sin\left( \pm \left( \pi - \arccos\frac{1}{4} + \pi n \right) + \pi n \right) = \frac{7}{2}\]

\[tg^{2}x = \frac{1}{5}\]

\[x = \pm arctg\frac{1}{\sqrt{5}} + \pi n.\]

\[tg^{2}x = \frac{2}{7}\]

\[x = \pm arctg\ \sqrt{\frac{2}{7}} + \pi n.\]

\[Ответ:\ \]

\[\left\{ \begin{matrix} x = \pm arctg\frac{1}{\sqrt{5}} + \pi n \\ y = \pm \arccos\frac{1}{4} + \pi n \\ \end{matrix} \right.\ \text{\ \ \ \ \ }или\ \ \ \ \ \ \]

\[\left\{ \begin{matrix} x = \pm arctg\ \sqrt{\frac{2}{7}} + \pi n\ \ \ \ \ \ \ \ \ \ \ \\ y = \pm \left( \pi - \arccos\frac{1}{8} \right) + \pi n \\ \end{matrix} \right.\ \]

\[2)\ \left\{ \begin{matrix} tg\ x + tg\ y + tg\ x\ tg\ y = 1 \\ \sin{2y} - \sqrt{2}\sin x = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Если\ \ y = - x + \frac{\pi}{4}:\]

\[tg\ x + tg\ \left( \frac{\pi}{4} - x \right) +\]

\[+ \text{tg\ x\ tg}\left( - x + \frac{\pi}{4} \right) + 1.\]

\[Для\ любого\ x:\]

\[y = - x + \frac{\pi}{4} + \pi n.\]

\[\sin\left( - 2x + \frac{\pi}{2} + 2\pi n \right) -\]

\[- \sqrt{2}\sin x = \cos{2x} -\]

\[- \sqrt{2}\sin x = 1\]

\[1 - 2\sin^{2}x - \sqrt{2}\sin x = 1\]

\[\sqrt{2}\sin^{2}x - \sin x = 0\]

\[\sin x = 0\]

\[x = \pi k.\]

\[\sqrt{2}\sin x = 1\]

\[\sin x = \frac{1}{\sqrt{2}}\]

\[x = ( - 1)^{n} \cdot \frac{\pi}{4} + \pi k\]

\[x = \frac{\pi}{4} + 2\pi k;\ \ \ x = \frac{3\pi}{4} + 2\pi k.\]

\[Ответ:\ \]

\[\textbf{а)}\ \left\{ \begin{matrix} x = \pi k\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y = - x + \frac{\pi}{4} + \pi n \\ \end{matrix} \right.\ \ \ \ \ \ \ \rightarrow\]

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

\[\textbf{б)}\ \left\{ \begin{matrix} x = \frac{\pi}{4} + 2\pi k\ \ \ \ \ \ \ \ \ \ \ \ \ \\ y = 2\pi + \pi n - 2\pi k\ \\ \end{matrix} \right.\ \]

\[\textbf{в)}\ \left\{ \begin{matrix} x = \frac{3\pi}{4} + 2\pi k\ \ \ \ \ \ \ \ \ \ \ \ \ \\ y = - \frac{\pi}{2} + \pi n - 2\pi k \\ \end{matrix} \right.\ \]

\[3)\ \left\{ \begin{matrix} \cos{2x} - 2\cos^{2}y + 2 = 0 \\ \cos x\sqrt{\cos y} = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\cos x = 0 \rightarrow x = \frac{\pi}{2} + \pi n.\]

\[\cos y = 0 \rightarrow y = \frac{\pi}{2} + \pi n.\]

\[\textbf{а)}\ x = \frac{\pi}{2} + \pi n:\]

\[\cos(\pi + 2\pi n) - 2\sin^{2}y + 2 =\]

\[= 1 - 2\sin^{2}y - \cos{2y} = 0\]

\[\cos{2y} = 0\]

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

\[y = \frac{\pi}{4} + \frac{\text{πk}}{2}.\]

\[Нам\ подходят:\]

\[y = \pm \frac{\pi}{4} + 2\pi k.\]

\[\textbf{б)}\ y = \frac{\pi}{2} + \pi k\]

\[\cos{2x} - 2\cos\left( \frac{\pi}{2} + \pi k \right) + 2 =\]

\[= \cos{2x} + 2 = 0\]

\[Нет\ решений.\]

\[Ответ:\ \]

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

\[4)\ \left\{ \begin{matrix} \sqrt{1 + \sin x\sin y} = \cos y \\ 2ctg\ x\sin y + \sqrt{3} = 0\ \ \ \\ \end{matrix} \right.\ \]

\[1 + \sin x\sin y = \cos^{2}y\]

\[1 - \cos^{2}y + \sin x\sin y = 0\]

\[\sin^{2}y + \sin x\sin y = 0\]

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

\[\sin y = 0\]

\[y = \pi n.\]

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

\[2\sin\frac{x + y}{2}\cos\frac{y - x}{2} = 0\]

\[\sin\frac{x + y}{2} = 0\]

\[\frac{x + y}{2} = \text{πn}\]

\[y = - x + 2\text{πn.}\]

\[\cos\frac{y - x}{2} = 0\]

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

\[y = x + \pi + 2\pi n.\]

\[Нам\ подходит\ корень\ y = 2\pi n:\]

\[2ctg\ x\sin{2\pi n} + \sqrt{3} = 0\]

\[1)\ \ \left\{ \begin{matrix} y = - x + 2\pi n\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2ctg\ x\sin y + \sqrt{3} = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[2)\ \left\{ \begin{matrix} y = x + \pi + 2\pi n\ \ \ \ \ \ \ \ \ \\ 2ctg\ x\sin y + \sqrt{3} = 0 \\ \end{matrix} \right.\ \]

\[0 + \sqrt{3} = 0\]

\[нет\ решений.\]

\[Первый\ случай:\]

\[- 2ctg\ x\sin x + \sqrt{3} = 0\]

\[- 2\cos x + \sqrt{3} = 0\]

\[\cos x = \frac{\sqrt{3}}{2}\]

\[x = \pm \frac{\pi}{6} + 2\pi k.\]

\[Второй\ случай:\]

\[- 2ctg\ x\sin x + \sqrt{3} = 0\]

\[x = \pm \frac{\pi}{6} + 2\pi k.\]

\[Нам\ подходят:\]

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

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