<s>[INST] Finding more critical points If we have the function $f$ defined: 
$$f(x,y)=2\sin(x)+2\sin(y)+\sin(x+y)$$ for $-\pi \le x\le \pi$ and $-\pi \le y \le \pi$
Find the critical points and determine the nature of each.
I'm a bit stuck on this.
I've found: 
$\frac {\partial f}{\partial x}=2\cos(x)+\cos(x+y)$
$\frac {\partial f}{\partial y}=2\cos(y)+\cos(x+y)$
$\frac {\partial^2 f}{\partial x^2}=-2\sin(x)-\sin(x+y)$
$\frac {\partial^2 f}{\partial y^2}=-2\sin(y)-\sin(x+y)$
$\frac {\partial^2 f}{\partial x\partial y}=-\sin(x+y)$
Solving $f_x=0$ and $f_y=0$ I get $x=\arccos\left(\frac{-1\pm\sqrt 3}{2}\right)$ which has only one real solution, but looking at the graph there should be more than one critical point in the region. How should I find the others?
 [/INST] Based on your comment the function is $f(x,y)=2\sin(x)+2\sin(y)+\sin(x+y)$, to solve critical points,
$f_x=2cos(x)+cos(x+y)=0$ and
$f_y=2cos(y)+cos(x+y)=0$
Subtract equations we have $\cos(x)=\cos(y)$, when $x$ and $y$ is constrained on given range, that means $x=y$ or $x=-y$.
When $x=y$, $2\cos(x)+\cos(2x)=0\Rightarrow 2\cos(x)+2\cos^2(x)-1=0\Rightarrow \cos(x)=\frac{-1+\sqrt{3}}{2}$ ($\frac{-1-\sqrt{3}}{2}<-1$ therefore outside the range of $\cos(x)$)
That means $x=y=\pm\arccos(\frac{-1+\sqrt{3}}{2})$.
When $x=-y$, $\cos(x+y)=1\Rightarrow 2cos(x)=-1\Rightarrow cos(x)=-\frac{1}{2}$
$x=\pm 2\pi/3$, $y=-x=\mp2\pi/3$.
So critical points are $(\arccos(\frac{-1+\sqrt{3}}{2}),\arccos(\frac{-1+\sqrt{3}}{2}))$, $(-\arccos(\frac{-1+\sqrt{3}}{2}), -\arccos(\frac{-1+\sqrt{3}}{2}))$, $(2\pi/3,-2\pi/3)$, $(-2\pi/3,2\pi/3)$
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