Arc?

Consider a right-angled triangle $ABC$ with $\angle C=90^{\circ}$ , $AC=\frac{x}{3}$ , $BC=\frac{x}{2}$ and $BC=1$ (the hypotenuse).

Then $\arcsin{\frac{x}{2}}=\arccos{\frac{x}{3}}=\angle A$ , and also (by Pythagoras), $\frac{x^2}{4}+\frac{x^2}{9}=1$ .

Solving this gives $x=\frac{6}{\sqrt{13}}=\boxed{1.664\ldots}$ .

Let $\sin^{-1} \dfrac x2 = \cos^{-1} \dfrac x3 = \theta$ . Then $\sin \theta = \dfrac x2$ and $\cos \theta = \dfrac x3$ and

$\begin{aligned} \sin^2 \theta + \cos^2 \theta & = 1 \\ \left(\frac x2\right)^2 + \left(\frac x3\right)^2 & = 1 \\ \frac {13}{36}x^2 & = 1 \\ \implies x & = \frac 6{\sqrt {13}} \approx \boxed{1.644} & \small \blue{\text{Taking the greatest value of }x} \end{aligned}$