ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1238

\[\boxed{\mathbf{1238}\mathbf{.}}\]

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

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

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

\[Пусть\ tgx = y:\]

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

\[D = 1 + 24 = 25\]

\[y_{1} = \frac{- 1 + 5}{6} = \frac{2}{3};\ \ \ \]

\[y_{2} = \frac{- 1 - 5}{6} = - 1.\]

\[1)\ tg\ x = - 1\]

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

\[2)\ tg\ x = \frac{2}{3}\]

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

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

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

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

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

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

\[Пусть\ tg\ x = y:\]

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

\[D = 9 + 16 = 25\]

\[y_{1} = \frac{- 3 - 5}{4} = - 2;\ \ \]

\[y_{2} = \frac{- 3 + 5}{4} = \frac{1}{2}.\]

\[1)\ tg\ x = - 2\]

\[x = - arctg\ 2 + \pi n.\]

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

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

\[Ответ:\ \ x = - arctg\ 2 + \pi n;\ \ \]

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