Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 736

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

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

\[( - \pi;\ 2\pi):\]

\[x_{1} = \frac{\pi}{4} - \pi = - \frac{3\pi}{4};\]

\[x_{2} = \frac{\pi}{4};\]

\[x_{3} = \frac{\pi}{4} + \pi = \frac{5\pi}{4}.\]

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

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

\[( - \pi;\ 2\pi):\]

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

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

\[x_{3} = \frac{\pi}{3} + \pi = \frac{4\pi}{3}.\]

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

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

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

\[( - \pi;\ 2\pi):\]

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

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

\[x_{3} = - \frac{\pi}{3} + 2\pi = \frac{5\pi}{3}.\]

\[4)\ tg\ x = - 1\ \]

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

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

\[( - \pi;\ 2\pi):\]

\[x_{1} = - \frac{\pi}{4};\]

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

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