Double trouble $\cos^{-1}$

Let $\cos a=\dfrac{2}{3x}$ and $\cos b =\dfrac{3}{4x}$

Then, $a+b=\dfrac{\pi}{2}$

$\cos b =\cos \bigg(\dfrac{\pi}{2} -a \bigg)$

$\cos b=\sin a$

$\sin a=\dfrac{3}{4x}$

$\cos a=\dfrac{\sqrt{16x^{2} -9}}{4x}$

Also from the equation , $\cos a=\dfrac{2}{3x}$

$\dfrac{2}{3x}=\dfrac{\sqrt{16x^{2} -9}}{4x}$

Solving for x gives two possible values but since we are looking for a number > $\dfrac{3}{4}$ , $x=\dfrac{\sqrt{145}}{12}$

$x=\dfrac{\sqrt{12^{2} +1}}{12}$

$\begin{aligned} \cos^{-1} \left(\frac 2{3x}\right) + \cos^{-1} \left(\frac 3{4x}\right) & = \frac \pi 2 \\ \left(\frac 2{3x}\right) \left(\frac 3{4x}\right) - \sqrt{\left(1-\left(\frac 2{3x}\right)^2\right)\left(1-\left(\frac 3{4x}\right)^2\right)} & = 0 \\ \sqrt{\left(1-\frac 4{9x^2}\right)\left(1-\frac 9{16x^2}\right)} & = \frac 1{2x^2} \\ \frac 1{12x^2}\sqrt{\left(9x^2-4\right)\left(16x^2 - 9\right)} & = \frac 1{2x^2} & \small \color{#3D99F6} \text{For }x \ne 0 \\ \sqrt{144x^4 - 145x^2 + 36} & = 6 & \small \color{#3D99F6} \text{Squaring both sides} \\ 144x^4 - 145x^2 + 36 & = 36 \\ x^2 \left(144x^2 - 145\right) & = 0 & \small \color{#3D99F6} \text{For }x > \frac 34 \\ \implies x & = \sqrt{\frac {144+1}{12}} \end{aligned}$

Therefore, $A=\boxed{12}$ .