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

Question

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?

Answer 1

Hint:$$\sin(\arccos(y))=-\frac{1}{2}\left(x+\frac{1}{x}\right)$$ and $$|\sin(x)|\le 1$$

Answer 2

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 $x=\pm1$

Answer 3

Hint:

The discriminant is $$(2\sqrt{1-y^2})^2-4\cdot1\cdot1=-4y^2$$

$$x=\dfrac{-2\sqrt{1-y^2}\pm2iy}2$$

$\implies$ for real $x,$

$y$ must be $0$