Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 641

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

\[y = \frac{\cos{2x}}{\cos x}:\]

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

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

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

\[D = 1^{2} - 4 = 1 - 4 = - 3 < 0\]

\[корней\ нет.\]

\[Ответ:\ \ корней\ нет\text{.\ }\]

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

\[y = \sin x:\]

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

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

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

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

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

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

\[\sin x = 1\]

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

\[Ответ:\ \ \frac{\pi}{2} + 2\pi n.\]