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I have the following system of equations:

\begin{cases} \frac{\cos (x)}{1+y^2}=0 \\ \frac{-2y\cdot \sin(x)}{(1+y^2)^2}=0 \end{cases}

The first equation has the solution $x=k\pi-\pi/2$ for an integer $k$ and any real $y$. The second one has the solutions $x=k\pi$ for any real $y$ OR $y=0$ for any real $x$. My question is: how does one find the solution(s) of the system of equation?

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    $\begingroup$ $y=0, x=k\pi-\pi/2$ follows from what you've said. What's the difficultly? $\endgroup$ – saulspatz Dec 25 '20 at 17:30
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Well, we are trying to solve the following system of equations:

$$ \begin{cases} \cos\left(x\right)=0\\ \\ \text{n}\sin\left(x\right)=0 \end{cases}\tag1 $$

It is not hard to see, from the first equation, that:

$$x=\pi\text{k}\pm\frac{\pi}{2}\tag2$$

Where $\text{k}\in\mathbb{Z}$.

Substituting that in the second equation, gives:

$$\text{n}\sin\left(\pi\text{k}\pm\frac{\pi}{2}\right)=0\tag3$$

We can simplify the LHS:

$$\text{n}\left(\mp1\right)=0\space\Longleftrightarrow\space\text{n}=0\tag4$$

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