Given this problem Restrictions on $x$ any are that $x\in[0,\pi]$ , $y\in[0,\pi]$
I have $f_x=-(\cos y)({\sin(2x+y))}--------*$
$f_y=-(\cos x)(\sin x+2y)-----------**$
So from $*$ I get either $\cos y=0$ or $\sin(2x+y)=0$
From $\sin(2x+y)=0$ , I get $2x+y=0,\pi,2\pi,3\pi$ as $2x+y\in[0,3\pi]$
And from second case I get $\cos y=0\implies y=\dfrac{\pi}{2}$
Also from $**$ , I get $x+2y=0,\pi,2\pi,3\pi$ and from $\cos x=0\implies x=\dfrac{\pi}{2}$
Now I have made $16$ cases because of following equations:
$2x+y=0,\pi,2\pi,3\pi$
$x+2y=0,\pi,2\pi,3\pi$
example as like $2x+y=0$ , $x+2y=3\pi$ ....similarly $16$ cases from above equations
and I point $\left(\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$
After examining all cases I have got valid critical points only for cases of type
$2x+y=R$
$x+2y=R$
Where $R$ is $0,\pi,2\pi,3\pi$ , taken one by one. So out of $16$ cases I have made above , I got my valid points out of $4$ cases as shown above..
MY QUESTION - Why I am getting points from these cases only ($R$ cases) ...Points from other $12$ cases are either same as that of the ($R$ cases), or they are not in domain
Also am I right in making $16$ cases like this. Is this correct way of solving these equations?
Is there any alternative way??
Thanks for your kind help!!
You have $f(x,y)=\cos x\cos y\cos(x+y)$, so $$ f'_x=-\sin x\cos y\cos(x+y)-\cos x\cos y\sin(x+y)=-\cos y\sin(2x+y) $$ Since interchanging $x$ and $y$ doesn't change the function, we have $$ f'_y=-\cos x\sin(x+2y) $$ We have four cases: $$ \begin{cases} \cos y=0\\ \cos x=0 \end{cases} \qquad \begin{cases} \cos y=0\\ \sin(x+2y)=0 \end{cases} \qquad \begin{cases} \sin(2x+y)=0\\ \cos x=0 \end{cases} \qquad \begin{cases} \sin(2x+y)=0\\ \sin(x+2y)=0 \end{cases} $$
The first case gives $x=y=\pi/2$.
The second case gives $y=\pi/2$ and $\sin(x+\pi)=0$, that is, $-\sin x=0$, so critical points at $(0,\pi/2)$ and $(\pi,\pi/2)$.
The third case is symmetric, giving critical points at $(\pi/2,0)$ and $(\pi/2,\pi)$.
Let's tackle the last case. From $\sin(2x+y)=0$ we get $2x+y=m\pi$ for some integer $m$; similarly, $x+2y=n\pi$ for an integer $n$. Now let's solve the linear system $$ \begin{cases} 2x+y=m\pi\\ x+2y=n\pi \end{cases} $$ that gives $$ \begin{cases} x=(2m-n)\dfrac{\pi}{3}\\[6px] y=(2n-m)\dfrac{\pi}{3} \end{cases} $$ With the given limitations we must have $0\le 2m-n\le 3$ and $0\le 2n-m\le 3$ and we should determine $m$ and $n$ so that this holds.
If we set $2m-n=h$ and $2n-m=k$, we get $$ m=\frac{2h+k}{3},\qquad n=\frac{h+2k}{3} $$ So we cannot have arbitrary choices for $h$ and $k$, since we want integer values of $m$ and $n$. The only values are so \begin{alignat}{3} &m=0, n=0&\quad&\text{corresponding to}&\quad& h=0, k=0 \\ &m=1, n=1,&&\text{corresponding to}&& h=1, k=1 \\ &m=1, n=2&&\text{corresponding to}&& h=0, k=3 \\ &m=2, n=1,&&\text{corresponding to}&& h=3, k=0 \\ &m=2, n=2,&&\text{corresponding to}&& h=2, k=2 \\ &m=3, n=3&&\text{corresponding to}&& h=3, k=3 \end{alignat} and the critical points are $$ \left(0,0\right) \qquad \left(\frac{\pi}{3},\frac{\pi}{3}\right) \qquad \left(0,\pi\right) \qquad \left(\pi,0\right) \qquad \left(\frac{2\pi}{3},\frac{2\pi}{3}\right) \qquad \left(\pi,\pi\right) $$
See the following diagram drawn with Geogebra to see why