ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 742

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\[1)\ tg\ 2x = \sqrt{3}\]

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

\[x = \frac{1}{2} \bullet \left( \frac{\pi}{3} + \pi n \right) = \frac{\pi}{6} + \frac{\text{πn}}{2}\]

\[\left( - \frac{\pi}{2};\ \pi \right):\]

\[x_{1} = \frac{\pi}{6} - \frac{\pi}{2} = - \frac{\pi}{3};\]

\[x_{2} = \frac{\pi}{6};\]

\[x_{3} = \frac{\pi}{6} + \frac{\pi}{2} = \frac{2\pi}{3}.\]

\[2)\ tg\ 3x = - 1\]

\[3x = - arctg\ 1 + \pi n\]

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

\[x = \frac{1}{3} \bullet \left( - \frac{\pi}{4} + \pi n \right)\]

\[x = - \frac{\pi}{12} + \frac{\text{πn}}{3}\]

\[\left( - \frac{\pi}{2};\ \pi \right):\]

\[x_{1} = - \frac{\pi}{12} - \frac{\pi}{3} = \frac{5\pi}{12};\]

\[x_{2} = - \frac{\pi}{12};\]

\[x_{3} = - \frac{\pi}{12} + \frac{\pi}{3} = \frac{\pi}{4};\]

\[x_{4} = - \frac{\pi}{12} + \frac{2\pi}{3} = \frac{7\pi}{12};\]

\[x_{5} = - \frac{\pi}{12} + \pi = \frac{11\pi}{12}.\]