How to show (x, y) = (0, 0) is the only solution to $4x^2 + 3y^2 + \cos(2x^2 + y^2) = 1$ I can't seem to  think of/find a solution to this problem:

Show that $(x, y) = (0, 0)$ is the only solution to $4{x}^{2} + 3{y}^2 + \cos(2x^{2}+y^{2}) = 1$ 

How would one go about proving this?
 A: Call $t=2x^2+y^2$. Clearly $t \ge 0$, and your equation can be rewritten as
$$2t+ \cos t + y^2=1$$
Now, it is easily checked that the function $2t+ \cos t$ is strictly increasing (simply compute its derivative, and check that it's $>0$), so that we have the following inequalities
$$1= 2t+ \cos t + y^2 \ge 2 \cdot 0 + \cos 0 + y^2 = 1 + y^2 \ge 1$$
which must be all equalities.
In particular, $y=0$ and $t=0$, hence $x=0$.
A: An alternate approach.
$$ F(x,y)=4{x}^{2} + 3{y}^2 + \cos(2x^{2}+y^{2}) - 1$$
is continuous and differentiable everywhere.
$$\dfrac{\partial F}{\partial x}=8x-4x\sin(2x^{2}+y^{2})=0 $$
implies that $x=0$ since $2-\sin(2x^{2}+y^{2})\ne0$.
Likewise
$$\dfrac{\partial F}{\partial y}=6y-2y\sin(2x^{2}+y^{2})=0 $$
implies that $y=0$ since $3-\sin(2x^{2}+y^{2})\ne0$.
So $(0,0)$ is the only critical point.
Application of the second derivative test will verify that $(0,0)$ is a minimum point of the function since
$F_{xx}(0,0)=8,F_{xy}(0,0)=F_{yx}(0,0)=0,F_{yy}(0,0)=6$ so $D=F_{xx}F_{yy}-F_{xy}^2>0$ and $F_{xx}>0$.
So the function has a unique minimum at $x=y=0$.
