Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 5

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

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

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

\[x \in R.\]

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

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

\[\cos x \geq 1;\]

\[x = 2\pi n.\]

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

\[\sin x > 0;\]

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

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

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

\[2\cos x \geq 1;\]

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

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

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

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

\[2\sin x \leq 1;\]

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

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

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

\[\cos x > 0;\]

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