Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 662

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\[1)\ tg^{2}\ x + 3\ tg\ x = 0\]

\[y = tg\ x:\]

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

\[y(y + 3) = 0\]

\[y_{1} = 0\ \ и\ \ y_{2} = - 3.\]

\[tg\ x = 0\]

\[x = arctg\ 0 + \pi n = \pi n.\]

\[tg\ x = - arctg\ 3 + \pi n.\]

\[Ответ:\ \ \pi n;\ \ - arctg\ 3 + \pi n.\]

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

\[y = tg\ x:\]

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

\[D = 1 + 24 = 25\]

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

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

\[tg\ x = - 1\]

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

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

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

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

\[Ответ:\ - \frac{\pi}{4} + \pi n;\ \ arctg\frac{3}{2} + \pi n.\]

\[3)\ tg\ x - 12\ ctg\ x + 1 = 0\]

\[tg\ x - \frac{12}{\text{tg\ x}} + 1 = 0\]

\[y = tg\ x:\]

\[y - \frac{12}{y} + 1 = 0\ \ \ \ \ | \bullet y\]

\[y^{2} + y - 12 = 0\]

\[D = 1 + 48 = 49\]

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

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

\[tg\ x = - 4\]

\[x = - arctg\ 4 + \pi n.\]

\[tg\ x = 3\]

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

\[Ответ:\ - arctg\ 4 + \pi n;\ \ \]

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

\[4)\ tg\ x + ctg\ x = 2\]

\[tg\ x + \frac{1}{\text{tg\ x}} - 2 = 0\]

\[tg^{2}\ x + \frac{1}{\text{tg\ x}} - 2 = 0\ \ \ \ \ | \bullet tg\ x\]

\[\text{tg}^{2}\ x - 2\ tg\ x + 1 = 0\]

\[(tg\ x - 1)^{2} = 0\]

\[tg\ x - 1 = 0\]

\[tg\ x = 1\]

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

\[Ответ:\ \ \frac{\pi}{4} + \pi n.\]