When does this equation $\cos(\alpha + \beta) = \cos(\alpha) + \cos(\beta)$ hold?

The following contour plots hint us why the third equation defies many attempts.

Another possible explanation is that, if we substitute $x = \tan(\alpha/2)$ and $y = \tan(\beta/2)$ then the formulas

$$ \sin \alpha = \frac{2x}{1+x^2}, \quad \cos \alpha = \frac{1-x^2}{1+x^2}, \quad \tan \alpha = \frac{2x}{1-x^2} $$

show that

\begin{align*} \sin(\alpha+\beta) = \sin\alpha+\sin\beta &\quad \Longleftrightarrow \quad xy(x+y) = 0, \\ \tan(\alpha+\beta) = \tan\alpha+\tan\beta &\quad \Longleftrightarrow \quad xy(x+y)(1-xy) = 0, \\ \cos(\alpha+\beta) = \cos\alpha+\cos\beta &\quad \Longleftrightarrow \quad 3x^2y^2 - x^2 - 4xy - y^2 - 1 = 0. \end{align*}

This may be another reason why our equation seems impossible to solve in simple terms. Finally, one interesting observation is that

$$ \cos(\alpha+\beta) + 1 = \cos\alpha + \cos\beta \quad \Longleftrightarrow \quad xy(1-xy) = 0 $$

can be easily solved.

Write $x=\cos\alpha$, $y=\cos\beta$, and now write $\cos(\alpha+\beta)$ in terms of just $x$ and $y$. Rearrange the terms of the equation $\cos(\alpha+\beta)=\cos\alpha+\cos\beta$ and square both sides. You should now get an equation you can use to solve for $y$ in terms of $x$. It doesn't look like the solution is going to be very pretty though!