Showing two equations have a solution for some $a$ For $a\in \Bbb R$ fixed, consider the following equations in $x,y\in\Bbb R$: 
$$ x + y + \sin(xy) = a \tag{1}$$ and $$\sin(x^2 + y) = 2a\tag{2}$$ 
I am asked to show that there exists $\epsilon>0$ such that these equations have a solution for every $a$ with $|a|< \epsilon$.
My attempt:
I think the solution $x = y = 0$ works, however I am unsure on how to choose which epsilon this is. From the second equation we have: $$|\sin(x^2 + y)| = 2|a| \leq 1$$ so would $\epsilon=1 /2$ work??
 A: I don't know what you think about these equations, but it seems obvious that we have to find a simultaneous solution $(x,y)$ to both equations, which then will depend on the parameter $a$.
To this end consider the map $f:\>{\mathbb R}^3\to{\mathbb R}^3$ given by
$$f:\quad (x,y,a)\mapsto\left\{\eqalign{u&:=x+y+\sin(xy)-a \cr v&:=\sin(x^2+y)-2a\cr w&:=a\cr}\right.$$
One easily computes $f({\bf 0})={\bf 0}$ and the Jacobian $J_f({\bf 0})=1$. It follows that $f$ maps a suitable neighborhood $U$ of ${\bf 0}\in{\mathbb R}^3$ diffeomorphically onto some neighborhood $V$ of ${\bf 0}$. So there is an inverse $C^1$-mapping $g:\>V\to U$, which has the form
$$g(u,v,w)=\bigl(\phi(u,v,w),\psi(u,v,w),w\bigr)\ .$$
The map
$$a\mapsto \pi_3\circ g(0,0, a)=\bigl(\phi(0,0,a),\psi(0,0,a)\bigr)$$
produces for each $a$ satisfying $|a|<\epsilon$ with a suitable $\epsilon>0$ a solution $(x,y)$ to the given system of equations. Here $\pi_3$ denotes the projection $(x,y,z)\mapsto (x,y).$
A: *

*Let $f(t)=2t+\sin(t^2)$, then $f$ is a continuous function and $\lim_{t\to \pm \infty}f(t)=\pm \infty$ which implies that $f$ is surjective. This implies that for every $a\in \Bbb R$ there exists $t\in \Bbb R$ such that $f(t)=a$. By setting $x=y=t$ we get that the first equation have a solution for every $a\in \Bbb R$.

*Let $g(t)=\sin(t)$, then clearly we have $g(\Bbb R)=[-1,1]$ and thus for every $a\in [-1/2,1/2]$, there exist $t\in \Bbb R$ such that $g(t)=2a$. By setting $x=0,y=t$ we get that the second equation have a solution for every $a\in [-1/2,1/2]$.
I leave you to conclude the exercise.
