ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 694

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\[1)\ y = \sqrt{\sin x + 1}\]

\[\sin x + 1 \geq 0\]

\[\sin x \geq - 1\]

\[при\ любом\ x.\]

\[Ответ:\ \ x \in R.\]

\[2)\ y = \sqrt{\cos x - 1}\]

\[\cos x - 1 \geq 0\]

\[\cos x \geq 1\]

\[\cos x = 1\]

\[x = \arccos 1 + 2\pi n = 2\pi n.\]

\[Ответ:\ \ x = 2\pi n.\]

\[3)\ y = \lg{\sin x}\]

\[\sin x > 0\]

\[\arcsin 0 + 2\pi n < x <\]

\[< \pi - \arcsin 0 + 2\pi n.\]

\[Ответ:\ \ 2\pi n < x < \pi + 2\pi n.\]

\[4)\ y = \sqrt{2\cos x - 1}\]

\[2\cos x - 1 \geq 0\]

\[2\cos x \geq 1\]

\[\cos x \geq \frac{1}{2}\]

\[- \arccos\frac{1}{2} + 2\pi n \leq x \leq\]

\[\leq \arccos\frac{1}{2} + 2\pi n.\]

\[Ответ:\ \]

\[- \frac{\pi}{3} + 2\pi n \leq x \leq \frac{\pi}{3} + 2\pi n.\]

\[5)\ y = \sqrt{1 - 2\sin x}\]

\[1 - 2\sin x \geq 0\]

\[- 2\sin x \geq - 1\]

\[\sin x \leq \frac{1}{2}\]

\[- \pi - \arcsin\frac{1}{2} + 2\pi n \leq x \leq \arcsin\frac{1}{2} + 2\pi n\]

\[- \pi - \frac{\pi}{6} + 2\pi n \leq x \leq \frac{\pi}{6} + 2\pi n.\]

\[Ответ:\ \]

\[- \frac{7\pi}{6} + 2\pi n \leq x \leq \frac{\pi}{6} + 2\pi n.\]

\[6)\ y = \ln{\cos x}\]

\[\cos x > 0\]

\[- \arccos 0 + 2\pi n < x < \arccos 0 + 2\pi n.\]

\[Ответ:\ \]

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