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МатематикаРешебник по алгебре и начала математического анализа 10 класс Алимов Задание 612
Задание 612
\[\boxed{\mathbf{612}\mathbf{.}}\]
\[1)\ (tg\ x - 1)\left( tg\ x + \sqrt{3} \right) = 0\]
\[1)\ tg\ x - 1 = 0\]
\[tg\ x = 1\]
\[x = arctg\ 1 + \pi n\]
\[x = \frac{\pi}{4} + \pi n\]
\[2)\ tg\ x + \sqrt{3} = 0\]
\[tg\ x = - \sqrt{3}\]
\[x = - arctg\ \sqrt{3} + \pi n\]
\[x = - \frac{\pi}{3} + \pi n\]
\[Ответ:\ \ x = \frac{\pi}{4} + \pi n;\ \]
\[x = \ - \frac{\pi}{3} + \pi n.\]
\[2)\ \left( \sqrt{3}\ tg\ x + 1 \right)\left( tg\ x - \sqrt{3} \right) = 0\]
\[1)\ \]
\[\sqrt{3}\ tg\ x + 1 = 0\]
\[\sqrt{3}\ tg\ x = - 1\]
\[tg\ x = - \frac{1}{\sqrt{3}}\]
\[x = - arctg\frac{1}{\sqrt{3}} + \pi n\]
\[x = - \frac{\pi}{6} + \pi n\]
\[2)\ tg\ x - \sqrt{3} = 0\]
\[tg\ x = \sqrt{3}\]
\[x = arctg\ \sqrt{3} + \pi n\]
\[x = \frac{\pi}{3} + \pi n\]
\[Ответ:\ x = - \frac{\pi}{6} + \pi n;\ \]
\[x = \ \frac{\pi}{3} + \pi n.\]
\[3)\ (tg\ x - 2)\left( 2\cos x - 1 \right) = 0\]
\[1)\ tg\ x - 2 = 0\]
\[tg\ x = 2\]
\[x = arctg\ 2 + \pi n\]
\[2)\ 2\cos x - 1 = 0\]
\[2\cos x = 1\]
\[\cos x = \frac{1}{2}\]
\[x = \pm \arccos\frac{1}{2} + 2\pi n\]
\[x = \pm \frac{\pi}{3} + 2\pi n\]
\[Ответ:\ \ x = arctg\ 2 + \pi n;\ \]
\[x = \ \pm \frac{\pi}{3} + 2\pi n.\]
\[4)\ (tg\ x - 4,5)\left( 1 + 2\sin x \right) = 0\]
\[1)\ tg\ x - 4,5 = 0\]
\[tg\ x = 4,5\]
\[x = arctg\ 4,5 + \pi n\]
\[2)\ 1 + 2\sin x = 0\]
\[2\sin x = - 1\]
\[\sin x = - \frac{1}{2}\]
\[x = ( - 1)^{n + 1} \bullet \arcsin\frac{1}{2} + \pi n\]
\[x = ( - 1)^{n + 1} \bullet \frac{\pi}{6} + \pi n\]
\[Ответ:\ \ x = arctg\ 4,5 + \pi n;\ \]
\[x = \ ( - 1)^{n + 1} \bullet \frac{\pi}{6} + \pi n.\]
\[5)\ (tg\ x + 4)\left( \text{tg}\frac{x}{2} - 1 \right) = 0\]
\[1)\ tg\ x + 4 = 0\]
\[tg\ x = - 4\]
\[x = - arctg\ 4 + \pi n\]
\[2)\ tg\frac{x}{2} - 1 = 0\]
\[\text{tg}\frac{x}{2} = 1\]
\[\frac{x}{2} = arctg\ 1 + \pi n = \frac{\pi}{4} + \pi n\]
\[x = 2 \bullet \left( \frac{\pi}{4} + \pi n \right)\]
\[x = \frac{\pi}{2} + 2\pi n\]
\[нет\ решения.\]
\[Ответ:\ x = - arctg\ 4 + \pi n.\]
\[6)\ \left( \text{tg}\frac{x}{6} + 1 \right)(tg\ x - 1) = 0\]
\[1)\ tg\frac{x}{6} + 1 = 0\]
\[\text{tg}\frac{x}{6} = - 1\]
\[\frac{x}{6} = - arctg\ 1 + \pi n = - \frac{\pi}{4} + \pi n\]
\[x = 6\left( - \frac{\pi}{4} + \pi n \right)\]
\[x = - \frac{3\pi}{2} + 6\pi n\]
\[нет\ корней.\]
\[2)\ tg\ x = 1\]
\[x = arctg\ 1 + \pi n\]
\[x = \frac{\pi}{4} + \pi n\]
\[Ответ:\ \ x = \frac{\pi}{4} + \pi n.\]
