Critical point of a function with $\sin$ and $\cos$

Question

I have to find and classify all the critical points of the function $f: K\to \mathbb{R}$

$$f(x,y) = \sin(2x)\cos(y)$$

inside the set $K = \{(x,y)\in\mathbb{R}^2 : 0\le x \le \pi , 0\le y \le 2\pi \}$

Well, K is bounded and closed, so it is compact. Thanks to Weierstrass, we know that $f$ has maximum and minimum in $K$. I calculated the partial derivatives, which are

$$ \frac{\partial f}{\partial x} = 2\cos(2x)\cos y $$ $$ \frac{\partial f}{\partial y} = -\sin(2x)\sin y $$

In the critical points che gradient is $0$, so we have to solve

$$ \begin{cases} \cos(2x)\cos y=0 \\ \sin(2x)\sin y=0 \end{cases} $$

$\cos(2x) = 0 \Leftrightarrow x = \pi/4, x = 3\pi/4$

$\cos(y) = 0 \Leftrightarrow y = \pi/2, y=3\pi/2$

$\sin(2x) = 0 \Leftrightarrow x = 0, x = \pi/2$

$\sin(y) = 0 \Leftrightarrow y = 0, y=\pi, y = 2\pi$

Is it possible that there are so many critic points?

Answer 1

These are the possible critical Points: $$\left(x=\frac{\pi }{4}\land (y=0\lor y=\pi \lor y=2 \pi )\right)\lor \left(x=\frac{3 \pi }{4}\land (y=0\lor y=\pi \lor y=2 \pi )\right)\lor \left(\left(x=0\lor x=\frac{\pi }{2}\lor x=\pi \right)\land \left(y=\frac{\pi }{2}\lor y=\frac{3 \pi }{2}\right)\right)$$

Answer 2

Careful! You have to solve these equations simultaneously. One cannot have both $\sin 2x=0$ and $\cos 2x=0$, and one cannot have both $\sin y=0$ and $\cos y=0$. To solve your system you have two possibilities: (i) $\sin 2x=0=\cos y$ and (ii) $\cos 2x=0=\sin y$. This will reduce the number of critical points you need to check.

Answer 3

First, I'll point out that you are missing one value in "$\cos(2x) = 0 \Leftrightarrow x = \pi/4$". Here, $x$ can be $\frac{ 3 \pi}{4}$ as well. Then you have to make sure that both derivatives have to be simultaneously zero. Making sure that your conditions (I, II, III and IV) are met as

(I or II) and (III or IV), there are 12 critical points in the domain.
$$(\frac{\pi}{4} \mbox{ or } \frac{3 \pi}{4}, 0)$$ $$(0 \mbox{ or } \frac{\pi}{2} \mbox{ or } \pi, \frac{\pi}{2})$$ $$( \frac{\pi}{4} \mbox{ or } \frac{3 \pi}{4}, \pi)$$ $$( 0 \mbox{ or } \frac{\pi}{2} \mbox{ or } \pi,\frac{3 \pi}{2})$$ $$(\frac{\pi}{4} \mbox{ or } \frac{3 \pi}{4}, 2\pi)$$