if ${\cos }^{-1}\left(\frac{2}{3x}\right)+{\cos }^{-1}\left(\frac{3}{4x}\right)=\frac{\pi }{2}\left(x>\frac{3}{4}\right)$ then $x$ is  equal to 

We have, ${\cos }^{-1}\left(\frac{2}{3x}\right)+{\cos }^{-1}\left(\frac{3}{4x}\right)=\frac{\pi }{2}$

$\begin{array}{l}\Rightarrow {\cos }^{-1}\left(\frac{3}{4x}\right)=\frac{\pi }{2}-{\cos }^{-1}\left(\frac{2}{3x}\right)\\ \Rightarrow {\cos }^{-1}\left(\frac{3}{4x}\right)={\sin }^{-1}\left(\frac{2}{3x}\right)\end{array}$

$\begin{array}{l}\Rightarrow \cos \left({\cos }^{-1}\left(\frac{3}{4x}\right)\right)=\cos \left({\sin }^{-1}\left(\frac{2}{3x}\right)\right)\\ \Rightarrow \frac{3}{4x}=\sqrt{1-\frac{4}{(3x{)}^2}}\Rightarrow \frac{3}{4}=\frac{\sqrt{9x^2-4}}{3}\end{array}$

Squaring both sides, we get

$\frac{9}{16}=\frac{9x^2-4}{9}\Rightarrow x^2=\frac{145}{144}\Rightarrow x=\frac{\sqrt{145}}{12}$