cos((x+T)^{2})=cos(x^{2})
\Rightarrow (x+T)^{2}= 2n\pi\;\pm x^{2}
\Rightarrow x^{2}+T^{2}+2Tx= 2n\pi\;\pm x^{2}
\Rightarrow T^{2}+2Tx= 2n\pi
OR
T^{2}+2Tx+ 2x^{2}= 2n\pi
in boh cases , we wont be able to find T independent of x , hence T isnt a constant ,
thus cos(x2) isnt periodic ,