Given equation

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

Let $\alpha=\cos^{-1}\left(\frac{2}{3x}\right), \qquad \beta=\cos^{-1}\left(\frac{3}{4x}\right)$ so that $\alpha+\beta=\frac{\pi}{2}.$

Use the complementary angle relation

Since $\alpha+\beta=\frac{\pi}{2}$, we have $\alpha=\frac{\pi}{2}-\beta.$ Taking cosine on both sides: $\cos\alpha=\cos\left(\frac{\pi}{2}-\beta\right)=\sin\beta.$

Now, $\cos\alpha=\frac{2}{3x}, \qquad \cos\beta=\frac{3}{4x}.$

Therefore, $\frac{2}{3x}=\sin\beta.$

Express $\sin\beta$ using $\cos\beta$

Since $\beta\in[0,\pi]$ and here $\alpha+\beta=\frac{\pi}{2}$ implies both are acute, so $\sin\beta>0$. Hence, $\sin\beta=\sqrt{1-\cos^2\beta}=\sqrt{1-\left(\frac{3}{4x}\right)^2}.$

Thus, $\frac{2}{3x}=\sqrt{1-\frac{9}{16x^2}}.$

Square both sides

$\frac{4}{9x^2}=1-\frac{9}{16x^2}.$

Move terms together: $\frac{4}{9x^2}+\frac{9}{16x^2}=1.$

Take LCM: $\frac{64+81}{144x^2}=1$ $\frac{145}{144x^2}=1.$

So, $144x^2=145$ $x^2=\frac{145}{144}$ $x=\frac{\sqrt{145}}{12}.$

Since $x>\frac{3}{4}>0$, we take the positive root.

Match with options

$x=\frac{\sqrt{145}}{12}$

This is Option C.

Verification with stored answer

Stored correct answer: C

Our derived answer: C

Hence, they agree.