Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 664

\[\boxed{\mathbf{664}\mathbf{.}}\]

\[1)\ 5\sin x + \cos x = 5\]

\[10\ tg\frac{x}{2} - 6\ tg^{2}\frac{x}{2} - 4 = 0\]

\[y = tg\frac{x}{2}:\]

\[10y - 6y^{2} - 4 = 0\]

\[3y^{2} - 5y + 2 = 0\]

\[D = 25 - 24 = 1\]

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

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

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

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

\[x = 2 \bullet \left( \text{arctg}\frac{2}{3} + \pi n \right)\]

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

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

\[\frac{x}{2} = arctg\ 1 + \pi n = \frac{\pi}{4} + \pi n\]

\[x = 2 \bullet \left( \frac{\pi}{4} + \pi n \right)\]

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

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

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

\[2)\ 4\sin x + 3\cos x = 6\]

\[8\ tg\frac{x}{2} - 9\ tg^{2}\frac{x}{2} - 3 = 0\]

\[y = tg\frac{x}{2}:\]

\[8y - 9y^{2} - 3 = 0\]

\[9y^{2} - 8y + 3 = 0\]

\[D = 64 - 108 = - 44 < 0\]

\[корней\ нет.\]

\[Ответ:\ \ корней\ нет.\]