# System of multivariable equations with trig functions

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?

• $y=0, x=k\pi-\pi/2$ follows from what you've said. What's the difficultly? – saulspatz Dec 25 '20 at 17:30

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:

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