I know that equation for d’Alembert’s equation is looking so: $g(x+y)+g(x-y)=2g(x)g(y)$. So I am trying to find actual solutions for this equation. First I took $x=y=0$ and I got $2g(0)=2g(0)^2$. From here $g(0)=0$ or $g(0)=1$. If I take $x=y$, then $g(2x)$+$g(0)=2g(x)^2$, and if we take $y=-x$, we will get $g(0)+g(2x)=2g(x)g(-x)$. 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 $\cos x$ with additional constant $a$ or $b$ as you would like. So the final form should be like this $g(x)=b+\cos x$, correct?