ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1215

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\[1)\sin\left( \frac{3\pi}{5} + x \right) = 2\sin\left( \frac{\pi}{5} - \frac{x}{2} \right)\]

\[\sin\left( \frac{\pi}{5} - \frac{x}{2} \right) -\]

\[- \sin\left( \frac{\pi}{2} - \left( \frac{3\pi}{10} + \frac{x}{2} \right) \right) =\]

\[= \cos\left( \frac{3\pi}{10} + \frac{x}{2} \right)\]

\[2\sin\left( \frac{3\pi}{10} + \frac{x}{2} \right)\cos\left( \frac{3\pi}{10} + \frac{x}{2} \right) =\]

\[= 2\cos\left( \frac{3\pi}{10} + \frac{x}{2} \right)\]

\[2\cos\left( \frac{3\pi}{10} + \frac{x}{2} \right) \cdot\]

\[\cdot \left( \sin\left( \frac{3\pi}{10} + \frac{x}{2} \right) - 1 \right) = 0\]

\[\cos\left( \frac{3\pi}{10} + \frac{x}{2} \right) = 0\]

\[\frac{3\pi}{10} + \frac{x}{2} = \frac{\pi}{2} + \pi n\]

\[\frac{x}{2} = \frac{\pi}{5} + \pi n\]

\[x = \frac{2\pi}{5} + 2\pi n.\]

\[\sin\left( \frac{3\pi}{10} + \frac{x}{2} \right) = 1\]

\[\frac{3\pi}{10} + \frac{x}{2} = \frac{\pi}{2} + \pi n\]

\[\frac{x}{2} = \frac{\pi}{5} + \pi n\]

\[x = \frac{2\pi}{5} + 2\pi n.\]

\[Ответ:x = \frac{2\pi}{5} + 2\pi n.\]

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

\[Пусть\ y = \sin x:\]

\[y + \frac{1}{y} = y^{2} + \frac{1}{y^{2}}\]

\[y^{3} + y = y^{4} + 1\]

\[y^{4} - y^{3} - y + 1 = 0\]

\[y^{3} \bullet (y - 1) - 1 \bullet (y - 1) = 0\]

\[\left( y^{3} - 1 \right)(y - 1) = 0\]

\[y_{1} = \sqrt[3]{1} = 1\ \ и\ \ y_{2} = 1\]

\[\sin x = 1\]

\[x = \arcsin 1 + 2\pi n = \frac{\pi}{2} + 2\pi n\]

\[Ответ:\ \ \frac{\pi}{2} + 2\pi n.\]

\[3)\cos^{2}{2x} + \frac{1}{\cos^{2}{2x}} =\]

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

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

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

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

\[\cos^{2}{2x} = t^{2} - 2:\]

\[t^{2} - 2 = t\]

\[t^{2} - t - 2 = 0\]

\[D = 1 + 8 = 9\]

\[t_{1} = \frac{1 + 3}{2} = 2;\ \ \ \]

\[t_{2} = \frac{1 - 3}{2} = - 1\]

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

\[Пусть\ y = \cos{2x}:\]

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

\[D = 1 - 4 < 0\ нет\ корней.\]

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

\[Пусть\cos{2x} = y:\]

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

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

\[y = 1.\]

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

\[2x = 2\pi n\]

\[x = \pi n.\]

\[Ответ:\ \ x = \pi n.\]