Решебник по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 20

\[\boxed{\mathbf{20.}}\]

\[\textbf{а)}\sin x(tg\ x - 1) = 0\]

\[1)\ \left\{ \begin{matrix} \sin x = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \pi k - \frac{\pi}{2} < x < \frac{\pi}{2} + \pi k \\ \end{matrix} \right.\ \]

\[\sin x = 0\]

\[x = \pi k.\]

\[2)\ \left\{ \begin{matrix} tg\ x - 1 = 0 \\ x \in R\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[tg\ x = 1\]

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

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

\[\textbf{б)}\ tgx\ \left( \sin x - 1 \right) = 0\]

\[1)\ \left\{ \begin{matrix} tg\ x = 0 \\ x \in R\ \ \ \ \\ \end{matrix} \right.\ \]

\[tg\ x = 0\]

\[x = \pi k.\]

\[2)\ \left\{ \begin{matrix} \sin x - 1 = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \pi n - \frac{\pi}{2} < x < \frac{\pi}{2} + \pi n \\ \end{matrix} \right.\ \]

\[\sin x = 1\]

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

\[нет\ решений.\]

\[Ответ:x = \pi k.\]

\[\textbf{в)}\ (tgx + 1)\cos x = 0\]

\[1)\ \left\{ \begin{matrix} tgx + 1 = 0 \\ x \in R\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

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

\[2)\ \left\{ \begin{matrix} \cos x = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \pi n - \frac{\pi}{2} < x < \frac{\pi}{2} + \pi n \\ \end{matrix} \right.\ \]

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

\[нет\ решений.\]

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

\[\textbf{г)}\ (ctgx - 1)\sin x = 0\ \]

\[1)\ \left\{ \begin{matrix} ctgx - 1 = 0 \\ x \in R\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[ctg\ x = 1\]

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

\[2)\ \left\{ \begin{matrix} \sin x = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \pi n < x < \pi + \pi n \\ \end{matrix} \right.\ \]

\[\sin x = 0\]

\[x = \pi n.\]

\[нет\ решений.\]

\[Ответ:x = \frac{\pi}{4} + \pi k.\]