Ok, I am currently trying to work on finding all real numbers $x$ and $y$ such that the equation
\begin{equation} \cos(x+y)=\cos x + \cos y \end{equation} holds. Here is what I have done so far. The LHS of the latter equation may be rewritten as \begin{equation} 2\cos^2\Big(\dfrac{x+y}{2}\Big)-1, \end{equation} while the RHS can be expressed as \begin{equation} 2\cos\Big(\dfrac{x+y}{2}\Big)\cos\Big(\dfrac{x-y}{2}\Big). \end{equation} This yields the equation \begin{equation} 2\cos^2\Big(\dfrac{x+y}{2}\Big)-1=2\cos\Big(\dfrac{x+y}{2}\Big)\cos\Big(\dfrac{x-y}{2}\Big) \end{equation} which is equivalent to \begin{equation} 2\cos^2\Big(\dfrac{x+y}{2}\Big)-2\cos\Big(\dfrac{x+y}{2}\Big)\cos\Big(\dfrac{x-y}{2}\Big)-1=0. \end{equation} Set $t=\cos\Big(\dfrac{x+y}{2}\Big)$. Then our equation becomes \begin{equation} 2t^2-2\cos\Big(\dfrac{x-y}{2}\Big)t-1=0. \end{equation} Solving for $t$ using the quadratic formula, one gets \begin{align} t&=\dfrac{2\cos(\frac{x-y}{2})\pm\sqrt{4\cos^2(\frac{x-y}{2})+8}}{4}\\&=\dfrac{\cos(\frac{x-y}{2})\pm\sqrt{\cos^2(\frac{x-y}{2})+2}}{2}. \end{align} I got stucked in that last step. Any suggestions about how to proceed will be appreciated a lot.
