Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 636

\[\boxed{\mathbf{636}\mathbf{.}}\]

\[4\ tg^{2}\ x - 5\ tg\ x - 6 = 0\]

\[y = tg\ x:\]

\[4y^{2} - 5y - 6 = 0\]

\[D = 25 + 96 = 121\]

\[y_{1} = \frac{5 - 11}{2 \bullet 4} = - \frac{6}{8} = - \frac{3}{4};\]

\[y_{2} = \frac{5 + 11}{2 \bullet 4} = 2.\]

\[1)\ tg\ x = - \frac{3}{4}\]

\[x = - arctg\frac{3}{4} + \pi n.\]

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

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

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

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

\[3\ tg^{2}\ x - 7\ tg\ x + 2 = 0\]

\[y = tg\ x:\]

\[3y^{2} - 7y + 2 = 0\]

\[D = 49 - 24 = 25\]

\[y_{1} = \frac{7 - 5}{2 \bullet 3} = \frac{2}{6} = \frac{1}{3};\]

\[y_{2} = \frac{7 + 5}{2 \bullet 3} = 2.\]

\[1)\ tg\ x = \frac{1}{3}\]

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

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

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

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

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

\[3)\ 1 - 4\sin x \bullet \cos x + 4\cos^{2}x = 0\]

\[tg^{2}\ x - 4\ tg\ x + 5 = 0\]

\[y = tg\ x:\]

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

\[D = 1 - 20 = - 19 < 0\]

\[корней\ нет.\]

\[Ответ:\ \ корней\ нет.\]

\[4)\ 1 + \sin^{2}x = 2\sin x \bullet \cos x\]

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

\[y = tg\ x:\]

\[2y^{2} - 2y + 1 = 0\]

\[D = 4 - 8 = - 4 < 0\]

\[корней\ нет.\]

\[Ответ:\ \ корней\ нет.\]