\[1)\ y = \sin x + \cos x\]
\[y(x + T) = y(x);\]
\[\sin(x + T) + \cos(x + T) =\]
\[= \sin x + \cos x\]
\[T_{1} = 2\pi;\]
\[T_{2} = 2\pi.\]
\[Ответ:\ \ 2\pi.\]
\[2)\ y = \sin x + tg\ x\]
\[y(x + T) = y(x);\]
\[\sin(x + T) + tg(x + T) =\]
\[= \sin x + tg\ x\]
\[T_{1} = 2\pi;\ \ \ T_{2} = \pi.\]
\[Ответ:\ \ 2\pi.\]
\[3)\ y = \sin x \bullet \sin{3x}\]
\[y = \frac{1}{2}\left( \cos{2x} - \cos{4x} \right)\]
\[y(x + T) = y(x);\]
\[\frac{\cos(2x + 2T) - \cos(4x + 4T)}{2} =\]
\[= \frac{\cos{2x} - \cos{4x}}{2}\]
\[\cos(2x + 2T) - \cos(4x + 4T) =\]
\[= \cos{2x} - \cos{4x}\]
\[2T_{1} = 2\pi;\ \ \ 4T_{2} = 2\pi;\]
\[T_{1} = \pi;\ \ \ \ \ \ \ \ \text{\ \ \ }T_{2} = \frac{\pi}{2}.\]
\[Ответ:\ \ \pi.\]
\[4)\ y = 2\ tg\frac{x}{2} - 3\ tg\frac{x}{3}\]
\[y(x + T) = y(x);\]
\[2\ tg\left( \frac{x}{2} + \frac{T}{2} \right) - 3\ tg\left( \frac{x}{3} + \frac{T}{3} \right) =\]
\[= 2\ tg\frac{x}{2} - 3\ tg\frac{x}{3}\]
\[\frac{T_{1}}{2} = \pi;\ \ \ \text{\ \ \ \ }\frac{T_{2}}{3} = \pi;\]
\[T_{1} = 2\pi;\ \text{\ \ \ }T_{2} = 3\pi.\]
\[Ответ:\ \ 6\pi.\]