How to find an implicit function from this relation?

Question

This is from Tenenbaum & Pollard's "Ordinary Differential Equations" book, chapter 1, Exercise 2, problem # 14:

Find the function g(x) that is implicitly defined by the relation: (1) $\sqrt{x^2 -y^2} + \arccos{(x/y)} = 0, y\neq0$

The answer is $y = g(x) = x$, and I have no idea how they got that.

My attempt:

$\sqrt{x^2 -y^2} + \arccos{(\frac{x}{y})} = 0$

$\arccos{(\frac{x}{y})} = -\sqrt{x^2 -y^2}$

$\cos(\arccos{(\frac{x}{y})}) = \cos(-\sqrt{x^2 -y^2})$

$\cos(\arccos{(\frac{x}{y})}) = \cos(\sqrt{x^2 -y^2})$

$\frac{x}{y} = \cos(\sqrt{x^2 -y^2})$

$y = \frac{x}{\cos(\sqrt{x^2 -y^2})}$

I'm not sure where to go after that...

Answer 1

For $\sqrt{x^2-y^2}\in \mathbb{R}$, we require $x^2\geq y^2$

For $\arccos(\frac xy)\in \mathbb{R}$, we require $|x|\leq|y|$

However $x=-y$ does not satisfy the equation. So the relation can only be satisfied by $x=y$

Answer 2

Since $\sqrt{x^2-y^2}$ is defined, we have $|y|\leq|x|$.
Since $\arccos(x/y)$ is defined, we have $|y|\geq|x|$.