Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1220

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

\[tg\ x + tg\ y = \frac{\sin(x + y)}{\cos x\cos y} =\]

\[= \sqrt{2} \cdot \sin(x + y) = 1\]

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

\[x + y = \frac{\pi}{4} + 2\pi n;\]

\[x + y = \frac{3\pi}{4} + 2\pi n.\]

\[\cos x\cos\left( \frac{\pi}{4} - x \right) = \frac{1}{\sqrt{2}}\]

\[\frac{\sqrt{2}}{2}\sin x\left( \cos x - \sin x \right) = 0\]

\[\sin x = 0\]

\[x = \pi k.\]

\[1 - tgx = 0\]

\[tgx = 1\]

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

\[\cos x\cos\left( \frac{3\pi}{4} - x \right) - \frac{1}{\sqrt{2}} = 0\]

\[- \frac{\sqrt{2}}{2}\left( \cos^{2}x - \sin x\cos x + 1 \right) =\]

\[= 0\]

\[2\cos^{2}x - \sin x\cos x +\]

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

\[tg^{2}x - tgx + 2 = 0\]

\[Пусть\ tg\ x = y:\]

\[y^{2} - y + 2 = 0\]

\[D = 1 - 8 =\]

\[= - 7 < 0\ (нет\ корней).\]

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

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

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

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

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

\[\sin x\sin y = \cos x\cos y\]

\[\sin x\sin y - \cos x\cos y = 0\]

\[\cos(x + y) = 0\]

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

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

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

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

\[\cdot \text{tg\ }\left( \frac{\pi}{2} - x \right) = \sin^{2}x +\]

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

\[\frac{- 2\cos^{4}x + \cos^{2}x - 1}{2\cos^{2}x - 2} = 0\]

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

\[Пусть\cos^{2}x = y:\]

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

\[D = 1 - 8 =\]

\[= - 7 < 0\ (нет\ корней).\]

\[\textbf{б)}\ x + y = \frac{3\pi}{2} + 2\pi n\]

\[\sin x\sin\left( \frac{3\pi}{2} - x \right) + ctg\ x \cdot\]

\[\cdot \text{tg\ }\left( \frac{3\pi}{2} - x \right) = ctg^{2}x -\]

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

\[\frac{2\cos^{4}x - 7\cos^{2}x + 3}{2\cos^{2}x - 2} = 0\]

\[2\cos^{4}x - 7\cos^{2}x + 3 = 0\]

\[Пусть\ \cos^{2}x = y:\]

\[2y^{2} - 7y + 3 = 0\]

\[D = 49 - 24 = 25\]

\[y_{1} = \frac{7 + 5}{4} = 3;\ \ \ \]

\[y_{2} = \frac{7 - 5}{4} = \frac{1}{2}.\]

\[\cos^{2}x = \frac{1}{2}\]

\[\frac{1 + 2\cos x}{2} = \frac{1}{2}\]

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

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

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

\[\left\{ \begin{matrix} x = \frac{\pi}{4} + \frac{\text{πk}}{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ } \\ y = \frac{3\pi}{2} - x + 2\pi n \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = \frac{\pi}{4} + \frac{\text{πk}}{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ y = \frac{5\pi}{4} - \frac{\text{πk}}{2} + 2\pi n \\ \end{matrix} \right.\ \ \]