i know that equation for d’Alembert’s equation. is looking so $g(x+y)+g(x-y)=2*g(x)*g(y)$ so am trying to find actual solution for this equation,first i took $x=y=0$ and i got $2*g(0)$=$2*g(0)^2$ from here $g(0)=0$ or $g(0)=1$ if i take $x=y$ ,then $g(2*x)$+$g(0)$=$2*g(x)^2$,or if we take $y=-x$,we will get $g(0)$+$g(2*x)$=$(2*g(x)*g(-x))$ ,so i think that $g(x)=g(-x)$ ,am i right?so $g(x)$ should be of the form $g(x)$=$(x^n)$ where $n$ is even,or $g(x)=cos(x)$ right?
EDITED: but because $x^n$ for x=0 never equal 1,so it should be $cos(x)$ width additional additive constant a or b as you would like so final form should be like this $g(x)$=$b+cos(x)$ correct?