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