Critical points of $f(x,y)=\sin(x)+\sin(y)-\sin(x+y)$ Domain: $0 \le x<\pi$ and $0 \le y<\pi$
After setting the gradient of $f(x,y)$ equal to zero, I obtained the following system:
$\cos(x)=\cos(x+y),$  $\cos(y)=\cos(x+y)$
I am not sure how to solve this system but what I tried was:
On the given domain, the first equation implies $x=x $ and $y=0$.
The second equation implies $y=y $ and $x=0$ so the system is consistent when $(x,y)=(0,0)$?
And thus $(0,0)$ is the only critical point in the given domain?
 A: Setting the gradient to zero yields
$$
\begin{cases}
\cos x - \cos(x+y)=0 \\[6px]
\cos y - \cos(x+y)=0
\end{cases}
$$
Therefore $\cos x=\cos y$. If we are interested in the whole plane, this means
$$
y=x+2k\pi
\qquad\text{or}\qquad
y=-x+2k\pi
$$
With the second set of solutions we get $x+y=2k\pi$, so $\cos(x+y)=1$ and we obtain $\cos x=1$, so 
$$
x=2a\pi \qquad y=2b\pi
$$
for arbitrary integers $a$ and $b$.
With the first set of solutions we get $x+y=2x+2k\pi$, so $\cos(x+y)=\cos x$ becomes
$$
\cos 2x=\cos x
$$
This reduces to $2x=x+2h\pi$ or $2x=-x+2h\pi$. In the first case we get solutions we have already found, so it remains to consider $3x=2h\pi$.
This gives
$$
x=\frac{2h\pi}{3}\qquad y=\frac{2h\pi}{3}+2k\pi \tag{*}
$$
again for arbitrary integers $h$ and $k$.
The only solutions that satisfy $0\le x<\pi$ and $0\le y<\pi$ are
$$
\begin{cases}
x=0\\[6px]
y=0
\end{cases}
\qquad
\begin{cases}
x=\dfrac{2\pi}{3}\\[6px]
y=\dfrac{2\pi}{3}
\end{cases}
$$

If you want the solutions in (*) to be “symmetric” in $x$ and $y$, you can write them in the forms
$$
\begin{cases}
x=\dfrac{2\pi}{3}+2a\pi \\[6px]
y=\dfrac{2\pi}{3}+2b\pi
\end{cases}
\quad
\begin{cases}
x=\dfrac{4\pi}{3}+2a\pi \\[6px]
y=\dfrac{4\pi}{3}+2b\pi
\end{cases}
$$
A: If both equations hold simultaneously, we'll have $\cos(x)=\cos(y)$. Since the cosine is injective in $[0,\pi]$, this implies $x=y$.
Now, we need only solve $\cos(x)=\cos(2x)$. Of course, $x=0$ is a solution, but are there more?
The double angle formula yields $\cos(2x)=2\cos(x)^2-1$, so we need to solve
$$2\cos(x)^2-\cos(x)-1=0$$
Letting $t=\cos(x)$, we have a quadratic equation $2t^2-t-1=0$. The solutions are
$$t=\frac{1\pm\sqrt{1+8}}{4}=\frac{1\pm3}{4}=1\,\text{ or }\, -\frac{1}{2}$$
The solution $t=1$ corresponds to $x=y=0$, while the solution $t=-\frac12$
 corresponds to $x=y=\frac{2\pi}3$, which is the critical point you missed.
