ГДЗ по алгебре и начала математического анализа 11 класс Колягин Проверь себя I

\[\mathbf{1.}\]

\[\mathbf{\ }y = tg\ 2x\]

\[2x \neq \frac{\pi}{2} + \pi n\]

\[x \neq \frac{\pi}{4} + \frac{\pi n}{2}.\]

\[y( - x) = tg( - 2x) =\]

\[= - tg\ 2x = - y(x).\]

\[Ответ:\ \ x \neq \frac{\pi}{4} + \frac{\pi n}{2};\ нечетная.\]

\[\mathbf{2.}\ \]

\[y = \sin x\ и\ y = \cos x:\]

\[1)\sin x = 1 \rightarrow x = - \frac{3\pi}{2};\]

\[\cos x = 1 \rightarrow x = - 2\pi;\]

\[2)\sin x = - 1 \rightarrow x \in \varnothing;\]

\[\cos x = 1 \rightarrow \ x = - \pi;\]

\[3)\sin x = 0 \rightarrow x = - 2\pi;\ x = - \pi;\]

\[\cos x = 0 \rightarrow x = - \frac{3\pi}{2};\]

\[4)\sin x > 0 \rightarrow - 2\pi < x < - \pi;\]

\[\cos x > 0 \rightarrow - 2\pi < x < - \frac{3\pi}{2};\]

\[5)\sin x < 0 \rightarrow \ x \in \varnothing;\]

\[\cos x < 0 \rightarrow - \frac{3\pi}{2} < x < - \pi.\]

\[\mathbf{3.}\]

\[\cos x < - \frac{1}{2};\ \ \ x \in \left\lbrack \frac{\pi}{2};\ 2\pi \right\rbrack:\]

\[\frac{2\pi}{3} + 2\pi n < x < \frac{4\pi}{3} + 2\pi n.\]

\[Ответ:\ \ \frac{2\pi}{3} < x < \frac{4\pi}{3}.\]

\[\mathbf{4.}\ \]

\[\text{ctg}\frac{\pi}{3};\ \text{ctg}\frac{7\pi}{8};\text{\ ctg}\frac{5\pi}{7};\text{\ ctg}\ 2.\]

\[y = ctg\ x\ убывает\ на\ (0;\ \pi):\]

\[0 < \frac{\pi}{3} < 2 < \frac{5\pi}{7} < \frac{7\pi}{8} < \pi.\]

\[Ответ:\ \ \]

\[\text{ctg}\frac{7\pi}{8},\ \ \ ctg\frac{5\pi}{7},\ \ \ ctg\ 2,\ \ \ ctg\frac{\pi}{3}.\]