ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 668

\[\boxed{\mathbf{668}\mathbf{.}}\]

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

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

\[y = tg\ x:\]

\[2y + 1 - 3y^{2} = 0\]

\[3y^{2} - 2y - 1 = 0\]

\[D = 4 + 12 = 16\]

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

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

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

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

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

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

\[Ответ:\ \]

\[- arctg\frac{1}{3} + \pi n;\ \ \frac{\pi}{4} + \pi n.\]

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

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

\[y = tg\ x:\]

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

\[2y^{2} - 3y + 1 = 0\]

\[D = 9 - 8 = 1\]

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

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

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

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

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

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

\[Ответ:\ \ \]

\[\text{arctg}\frac{1}{2} + \pi n;\ \ \frac{\pi}{4} + \pi n.\]