Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 659

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

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

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

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

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

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

\[x = - \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

\[Ответ:\ - \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

\[2)\ tg\left( 3x - \frac{\pi}{4} \right) = \frac{1}{\sqrt{3}}\]

\[3x - \frac{\pi}{4} = arctg\frac{1}{\sqrt{3}} + \pi n\]

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

\[3x = \frac{\pi}{6} + \frac{\pi}{4} + \pi n\]

\[3x = \frac{2\pi}{12} + \frac{3\pi}{12} + \pi n\]

\[3x = \frac{5\pi}{12} + \pi n\]

\[x = \frac{1}{3} \bullet \left( \frac{5\pi}{12} + \pi n \right)\]

\[x = \frac{5\pi}{36} + \frac{\text{πn}}{3}.\]

\[Ответ:\ \ \frac{5\pi}{36} + \frac{\text{πn}}{3}.\]

\[3)\ \sqrt{3} - tg\left( x - \frac{\pi}{5} \right) = 0\]

\[\text{tg}\left( x - \frac{\pi}{5} \right) = \sqrt{3}\]

\[x - \frac{\pi}{5} = arctg\ \sqrt{3} + \pi n\]

\[x - \frac{\pi}{5} = \frac{\pi}{3} + \pi n\]

\[x = \frac{\pi}{3} + \frac{\pi}{5} + \pi n\]

\[x = \frac{5\pi}{15} + \frac{3\pi}{15} + \pi n\]

\[x = \frac{8\pi}{15} + \pi n.\]

\[Ответ:\ \ \frac{8\pi}{15} + \pi n.\]

\[4)\ 1 - tg\left( x + \frac{\pi}{7} \right) = 0\]

\[\text{tg}\left( x + \frac{\pi}{7} \right) = 1\]

\[x + \frac{\pi}{7} = arctg\ 1 + \pi n\]

\[x + \frac{\pi}{7} = \frac{\pi}{4} + \pi n\]

\[x = \frac{\pi}{4} - \frac{\pi}{7} + \pi n\]

\[x = \frac{7\pi}{28} - \frac{4\pi}{28} + \pi n\]

\[x + \frac{\pi}{7} = \frac{3\pi}{28} + \pi n.\]

\[Ответ:\ \ \frac{3\pi}{28} + \pi n.\]