Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 1019

\[1)\ y = \cos{3x}\]

\[y(x + T) = y(x)\]

\[\cos{3(x + T)} = \cos{3x}\]

\[\cos(3x + 3T) = \cos{3x}\]

\[3T = 2\pi\]

\[T = \frac{2\pi}{3}.\]

\[Ответ:\ \ \frac{2\pi}{3}.\]

\[2)\ y = \sin\frac{x}{5}\]

\[y(x + T) = y(x)\]

\[\sin\frac{x + T}{5} = \sin\frac{x}{5}\]

\[\sin\left( \frac{x}{5} + \frac{T}{5} \right) = \sin\frac{x}{5}\]

\[\frac{T}{5} = 2\pi\]

\[T = 10\pi.\]

\[Ответ:\ \ 10\pi.\]

\[3)\ y = tg\ 5x\]

\[y(x + T) = y(x)\]

\[tg\ 5(x + T) = tg\ 5x\]

\[\text{tg}(5x + 5T) = tg\ 5x\]

\[5T = \pi\]

\[T = \frac{\pi}{5}.\]

\[Ответ:\ \ \frac{\pi}{5}.\]

\[4)\ y = \sin x + tg\ x\]

\[y(x + T) = y(x)\]

\[\sin(x + T) + tg(x + T) = \sin x + tg\ x\]

\[T = 2\pi,\ \ \ T = \pi\]

\[T = 2\pi.\]

\[Ответ:\ \ 2\pi.\]