Finding, the function $f(x)$ when $$f(0)=1 ~\text{and}~ f(10-x)=f(x), f(2-x)=f(2+x), \forall x \in R.......(1)$$
Let $x \to 2-x$ in $f(10-x)=f(x) \implies f(8+x)=f(2-x)=f(2+x)$
Next, we take $x\to x-2$ in above to get $f(6+x)=f(x), \forall x \in R$, implying that $f(x)$ is periodic with period $6$.
However, $f(x)=\cos(\pi x)$ does satisfy all of $(1)$ but $f(x)=\cos(\pi x/3)$ fails to do so. Where/What is the discrepancy?