ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 612

\[\boxed{\mathbf{612}\mathbf{.}}\]

\[1)\ (tg\ x - 1)\left( tg\ x + \sqrt{3} \right) = 0\]

\[1)\ tg\ x - 1 = 0\]

\[tg\ x = 1\]

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

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

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

\[tg\ x = - \sqrt{3}\]

\[x = - arctg\ \sqrt{3} + \pi n\]

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

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

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

\[2)\ \left( \sqrt{3}\ tg\ x + 1 \right)\left( tg\ x - \sqrt{3} \right) = 0\]

\[1)\ \]

\[\sqrt{3}\ tg\ x + 1 = 0\]

\[\sqrt{3}\ tg\ x = - 1\]

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

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

\[x = - \frac{\pi}{6} + \pi n\]

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

\[tg\ x = \sqrt{3}\]

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

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

\[Ответ:\ x = - \frac{\pi}{6} + \pi n;\ \]

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

\[3)\ (tg\ x - 2)\left( 2\cos x - 1 \right) = 0\]

\[1)\ tg\ x - 2 = 0\]

\[tg\ x = 2\]

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

\[2)\ 2\cos x - 1 = 0\]

\[2\cos x = 1\]

\[\cos x = \frac{1}{2}\]

\[x = \pm \arccos\frac{1}{2} + 2\pi n\]

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

\[Ответ:\ \ x = arctg\ 2 + \pi n;\ \]

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

\[4)\ (tg\ x - 4,5)\left( 1 + 2\sin x \right) = 0\]

\[1)\ tg\ x - 4,5 = 0\]

\[tg\ x = 4,5\]

\[x = arctg\ 4,5 + \pi n\]

\[2)\ 1 + 2\sin x = 0\]

\[2\sin x = - 1\]

\[\sin x = - \frac{1}{2}\]

\[x = ( - 1)^{n + 1} \bullet \arcsin\frac{1}{2} + \pi n\]

\[x = ( - 1)^{n + 1} \bullet \frac{\pi}{6} + \pi n\]

\[Ответ:\ \ x = arctg\ 4,5 + \pi n;\ \]

\[x = \ ( - 1)^{n + 1} \bullet \frac{\pi}{6} + \pi n.\]

\[5)\ (tg\ x + 4)\left( \text{tg}\frac{x}{2} - 1 \right) = 0\]

\[1)\ tg\ x + 4 = 0\]

\[tg\ x = - 4\]

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

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

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

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

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

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

\[нет\ решения.\]

\[Ответ:\ x = - arctg\ 4 + \pi n.\]

\[6)\ \left( \text{tg}\frac{x}{6} + 1 \right)(tg\ x - 1) = 0\]

\[1)\ tg\frac{x}{6} + 1 = 0\]

\[\text{tg}\frac{x}{6} = - 1\]

\[\frac{x}{6} = - arctg\ 1 + \pi n = - \frac{\pi}{4} + \pi n\]

\[x = 6\left( - \frac{\pi}{4} + \pi n \right)\]

\[x = - \frac{3\pi}{2} + 6\pi n\]

\[нет\ корней.\]

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

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

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

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