Calculation of total number of real ordered pairs $(x,y)$ 
Calculation of total number of real ordered pairs $(x,y)$
in $x^2-4x+2=\sin^2 y$ and $x^2+y^2\leq 3$

$\bf{My\; Try::}$ Given $x^2-4x+2=x^2-4x+4-2=\sin^2 y\Rightarrow (x-2)^2-2=\sin^2 y$
Now Using $0 \leq \sin^2 y\leq 1$. So we get $0\leq (x-2)^2-2\leq 1\Rightarrow 2\leq (x-2)^2\leq 3$
So we get $\sqrt{2}\leq |x-2|\leq \sqrt{3}\Leftrightarrow \bigg(-\sqrt{3}\leq (x-2)\leq \sqrt{3}\bigg)\;\;\cap\;\;\bigg((x-2)\leq -\sqrt{2} \;\cup\; (x-2)\;\geq \sqrt{2}\bigg)$
So we get $$x\in \left[2-\sqrt{3}\;\;,2-\sqrt{2}\right]\;\cup \;\left[2+\sqrt{2}\;,2+\sqrt{3}\right]$$
Now How can I solve it after that, Help me, Thanks


Now How can I solve it after that

I don't know how to continue from that. So, let us take another approach.
Solving $x^2-4x+2-\sin^2 y=0$ for $x$ gives
$$x=2\pm\sqrt{2+\sin^2y}$$
Now $x=2+\sqrt{2+\sin^2y}$ does not satisfy $x^2\le 3$. So, we have $x=2-\sqrt{2+\sin^2y}$.
Now
$$\left(2-\sqrt{2+\sin^2y}\right)^2+y^2\le 3$$
is equivalent to
$$3+\sin^2y+y^2\le 4\sqrt{2+\sin^2y}\tag1$$
By the way, for $-\pi/4\le y\le \pi/4$, we have
$$y^2\le \frac{\pi^2}{16},\quad 0\le \sin^2y\le \frac 12$$
and so using $\pi\le 4$ gives
$$3+\sin^2y+y^2\le 3+\frac 12+\frac{\pi^2}{16}\le 3+\frac 12+\frac{4^2}{16}=4\times 1.125\le 4\sqrt 2\le 4\sqrt{2+\sin^2y}$$
Hence, if $-\pi/4\le y\le \pi/4$, then $(1)$ holds.
It follows from this that there are infinitely many such pairs.
