ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1200

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

\[= 2\cos^{3}x\ \]

\[\frac{\sin^{3}x}{\cos^{3}x} + \frac{\sin^{2}x\cos x}{\cos^{3}x} -\]

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

\[tg^{3}\ x + tg^{2}x - 2 = 0\]

\[Пусть\ tg\ x = y:\]

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

\[P(1) = 0;\ \ y = 1 - корень.\]

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

\[y = 1;\ \ \ \ \]

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

\[D = 1 - 8 =\]

\[= - 7 < 0\ (нет\ корней).\]

\[tg\ x = 1\]

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

\[Ответ:\ \frac{\pi}{4} + \pi k.\]

\[2)\sin^{3}x - \sin^{2}x\cos x +\]

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

\[\frac{\sin^{3}x}{\cos^{3}x} - \frac{\sin^{2}x\cos x}{\cos^{3}x} +\]

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

\[tg^{3}\ x - tg^{2}x + 2 = 0\]

\[Пусть\ tg\ x = y:\]

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

\[P( - 1) = 0;\ \ y = - 1 - корень.\]

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

\[= (y + 1)\left( y^{2} - y \right) + 2\]

\[y = - 1.\]

\[tg\ x = - 1\]

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

\[Ответ:\ \ - \frac{\pi}{4} + \pi k.\]