ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1196

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\[1)\cos x = \sin x\ \ \ \ \ |\ :\cos x\ \]

\[1 = tg\ x\]

\[x = arctg\ 1 + \pi n = \frac{\pi}{4} + \pi n\]

\[Ответ:\ \ \frac{\pi}{4} + \pi n.\]

\[2)\sin x + \cos x = 0\ \ \ \ \ |\ :\cos x\]

\[tg\ x + 1 = 0\]

\[tgx = - 1\]

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

\[Ответ:\ - \frac{\pi}{4} + \pi n.\]

\[3)\ 2\sin^{2}x - 5\sin x\cos x -\]

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

\[2tg^{2}x - 5tgx - 3 = 0\]

\[Пусть\ tgx = y:\]

\[2y^{2} - 5y - 3 = 0\]

\[D = 25 + 24 = 49\]

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

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

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

\[x = - arctg\frac{1}{2} + \pi k.\]

\[2)\ tg\ x = 3\]

\[x = arctg\ 3 + \pi k.\]

\[Ответ:\ - arctg\frac{1}{2} + \pi k;\]

\[\ arctg\ 3 + \pi k.\]

\[4)\ 3\sin^{2}x - 14\sin x\cos x -\]

\[- 5\cos^{2}x = 0\ \ \ \ \ |\ :\cos^{2}x\ \]

\[3tg^{2}x - 14tg\ x - 5 = 0\]

\[Пусть\ tg\ x = y:\]

\[3y^{2} - 14y - 5 = 0\]

\[D_{1} = 49 + 15 = 64\]

\[y_{1} = \frac{7 + 8}{3} = 5;\ \ \]

\[\ y_{2} = \frac{7 - 8}{3} = - \frac{1}{3}.\]

\[1)\ tgx = - \frac{1}{3}\]

\[x = - arctg\frac{1}{3} + \pi k.\]

\[2)\ tg\ x = 5\]

\[x = arctg\ 5 + \pi k.\]

\[Ответ:\ - arctg\frac{1}{3} + \pi k;\ \]

\[\ arctg\ 5 + \pi k.\]