Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 686

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\[1)\ \left\{ \begin{matrix} \frac{\sin x}{\sin y} = \frac{5}{3} \\ \frac{\cos x}{\cos y} = \frac{1}{3} \\ \end{matrix} \right.\ ( + )\]

\[\frac{\sin x}{\sin y} + \frac{\cos x}{\cos y} = \frac{5}{3} + \frac{1}{3}\]

\[\frac{\sin x \bullet \cos y + \sin y \bullet \cos x}{\sin y \bullet \cos y} = \frac{6}{3}\]

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

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

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

\[\sin(x + y) - \sin{2y} = 0\]

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

\[\sin\frac{x - y}{2} \bullet \cos\frac{x + 3y}{2} = 0\]

\[1)\ \sin\frac{x - y}{2} = 0\]

\[\frac{x - y}{2} = \arcsin 0 + \pi n = \pi n\]

\[x - y = 2\pi n\]

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

\[2)\ \cos\frac{x + 3y}{2} = 0\]

\[\frac{x + 3y}{2} = \arccos 0 + \pi n =\]

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

\[x + 3y = \pi + 2\pi n\]

\[x = \pi + 2\pi n - 3y.\]

\[Подставим\ в\ первое\ уравнение:\]

\[\frac{\sin(y + 2\pi n)}{\sin y} = \frac{\sin y}{\sin y} =\]

\[= 1 \neq \frac{5}{3} - не\ подходит\]

\[\frac{\sin(\pi + 2\pi n - 3y)}{\sin y} = \frac{5}{3}\]

\[\frac{\sin{(\pi - 3y)}}{\sin y} = \frac{5}{3}\]

\[\frac{\sin{3y}}{\sin y} = \frac{5}{3}.\]

\[Формула\ тройного\ угла:\]

\[\frac{3\sin y - 4\sin^{3}y}{\sin y} = \frac{5}{3}\]

\[3 - 4\sin^{2}y = \frac{5}{3}\]

\[4\sin^{2}y = 3 - \frac{5}{3}\]

\[4\sin^{2}y = \frac{4}{3}\]

\[\sin^{2}y = \frac{1}{3}\]

\[\sin y = \pm \frac{1}{\sqrt{3}}\]

\[y_{1} = ( - 1)^{n + 1} \bullet \arcsin\frac{1}{\sqrt{3}} + \pi k;\]

\[y_{2} = ( - 1)^{n} \bullet \arcsin\frac{1}{\sqrt{3}} + \pi k;\]

\[y = \pm \arcsin\frac{1}{\sqrt{3}} + \pi k.\]

\[x = \pi \pm 3\arcsin\frac{1}{\sqrt{3}} + \pi(2n - k).\]

\[Ответ:\ \ \]

\[x = \pi \pm 3\arcsin\frac{1}{\sqrt{3}} + \pi(2n - k);\ \ \]

\[y = \pm \arcsin\frac{1}{\sqrt{3}} + \pi k.\]

\[2)\ \left\{ \begin{matrix} \sin x \bullet \cos y = \frac{1}{2}\text{\ \ \ \ } \\ \cos x \bullet \sin y = - \frac{1}{2} \\ \end{matrix} \right.\ ( + )\]

\[\sin x \bullet \cos y + \cos x \bullet \sin y = \frac{1}{2} - \frac{1}{2}\]

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

\[x + y = \arcsin 0 + \pi n = \pi n\]

\[x = \pi n - y.\]

\[Вычтем\ из\ первого\ второе:\]

\[\sin x \bullet \cos y - \cos x \bullet \sin y = \frac{1}{2} + \frac{1}{2}\]

\[\sin(x - y) = 1\]

\[x - y = \arcsin 1 + 2\pi k = \frac{\pi}{2} + 2\pi k\]

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

\[Получим:\]

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

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

\[y = - \frac{\pi}{4} - \pi k + \frac{\text{πn}}{2};\]

\[x = \pi n - y =\]

\[= \pi n + \frac{\pi}{4} + \pi k - \frac{\text{πn}}{2} =\]

\[= \frac{\pi}{4} + \pi k + \frac{\text{πn}}{2}.\]

\[Ответ:\ \ x = \frac{\pi}{4} + \pi k + \frac{\text{πn}}{2};\ \ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ }y = - \frac{\pi}{4} - \pi k + \frac{\text{πn}}{2}.\]