Solving $2y=\sqrt{3+\frac{1}{2y}}$

Question

Any way to solve this irrational equation in $\mathbb{R}$? I think it has some artifice, but I do not see it $$2y=\sqrt{3+\frac{1}{2y}}$$

Answer 1

Square it, clear the fraction, and you have a cubic. $$8y^3=6y+1$$ Alpha finds three real roots with complicated expressions, about $-0.76604, -0.17365,0.93969$.

Answer 2

Since $y$ equals to a square root, $y$ is non-negative.

Notice $$2y = \sqrt{3+\frac{1}{2y}} \implies 4y^2 = 3 + \frac{1}{2y} \implies 4y^3 - 3y = \frac12 $$ and for any $x \ge 1$, $4x^3 - 3x \ge (4x^2 - 3)x \ge 1$, we find $y \in (0,1)$.

Take a $\theta \in (0,\frac{\pi}{2})$ such that $y = \cos\theta$ and recall $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$. Last equation becomes

$$\cos(3\theta) = \cos\frac{\pi}{3}\quad\implies\quad 3\theta = (2N \pm \frac13)\pi\quad\text{ for some integer N} $$

Since $\theta \in (0,\frac{\pi}{2})$, this forces $\theta = \frac{\pi}{9}$ and hence $y = \cos\frac{\pi}{9}$