Решебник по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 39

\[\boxed{\mathbf{39.}}\]

\[\textbf{а)}\log_{0,5}{tg\ x + \left( \frac{1}{3} \right)^{\text{tgx}} - \sqrt[7]{\text{tg\ x}}} =\]

\[= \log_{0,5}\text{ctgx} + \left( \frac{1}{3} \right)^{\text{ctg\ x}} - \sqrt[7]{\text{ctg\ x}}\]

\[\left\{ \begin{matrix} tg\ x = ctg\ x \\ tg\ x > 0\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[tg\ x = ctg\ x\]

\[\frac{\sin x}{\cos x} = \frac{\cos x}{\sin x}\]

\[\text{si}n^{2}x = cos^{2}x\]

\[\text{co}s^{2}x - sin^{2}x = 0\]

\[\cos{2x} = 0\]

\[2x = \frac{\pi}{2} + \pi k\]

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

\[tg\ x > 0:\]

\[\pi k < x < \frac{\pi}{2} + \pi k.\]

\[Ответ:x = \frac{\pi}{4} + \pi k.\]

\[\textbf{б)}\log_{0,2}{\sin x} - 5^{\sin x} - \sqrt[4]{\sin x} =\]

\[\left\{ \begin{matrix} \sin x = - \cos x \\ \sin x > 0\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\sin x = - \cos x\]

\[\frac{\sin x}{\cos x} = - \frac{\cos x}{\sin x}\]

\[tg\ x = - 1\]

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

\[\sin x > 0:\]

\[2\pi k < x < \pi + 2\pi k.\]

\[Ответ:x = \frac{3\pi}{4} + 2\pi k.\]