Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 669

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

\[y = tg\ x:\]

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

\[D = 1 + 24 = 25\]

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

\[y_{2} = \frac{- 1 + 5}{2 \bullet 3} = \frac{4}{6} = \frac{2}{3}.\]

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

\[x = - arctg\ 1 + \pi n\]

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

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

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

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

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

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

\[y = tg\ x:\]

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

\[D = 9 + 16 = 25\]

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

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

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

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

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

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

\[Ответ:\ - arctg\ 2 + \pi n;\ \ \]

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