<s>[INST] Solve $x,y$ given $1+x^2+2x\sin(\arccos y)=0$ I have to solve $x,y$ given-

$$1+x^2+2x\sin(\arccos y)=0$$

My attempt:
$$1+x^2+2x\sin(\arcsin \sqrt{1-y^2})=0$$ (Is this step valid? Can I convert arccos to arcsin like this?)
$$\implies1+x^2+2x\sqrt{1-y^2}=0$$
Now,I can't procced further. What to do next?
 [/INST] Solving for x and y, $$1+x^2+2x\sin(\arccos y)=0$$
Firstly, if we take $z=\arccos(y)$ then $y=\cos(z)$ and so $\cos^2(z)=y^2$
we know that $\sin^2(z)+\cos^2(z)=1$ so 
$1-\sin^2(z)=y^2\,\therefore \sin(z)=\sqrt{1-y^2}\therefore z=\arcsin(\sqrt{1-y^2})$
If we put this back into our original expression we get:
$$1+x^2+2x\sin\left(\arcsin\left(\sqrt{1-y^2}\right)\right)=0$$
so:
$$1+x^2+2x\sqrt{1-y^2}=0$$
this can be rearranged to give:
$$\sqrt{1-y^2}=-\frac{1+x^2}{2x}$$
This is the same as what you put, so it is correct so far. Now I would rearrange to:
$$y=\sqrt{1-\left(\frac{1+x^2}{2x}\right)^2}$$
For $y=0$,
$$\frac{1+x^2}{2x}=\pm1$$
solving this we get two equations, $x^2+2x+1=0$ and $x^2-2x+1=0$
these can be factorised to $(x+1)^2=0$ and $(x-1)^2=0$ giving the two clear solutions of $x=1$ and $x=-1$
So the answer is $x=\pm1$
</s>