The value of x, if $cos^{-1}\sqrt{3}x+cos^{-1}x=\frac{\pi}{2}$ is :

$\frac{1}{2}$

Equation: $\cos^{-1}(\sqrt{3}x)+\cos^{-1}(x)=\frac{\pi}{2}$

Set $u=\sqrt{3}x,\;v=x$. Then $\cos^{-1}u+\cos^{-1}v=\frac{\pi}{2}$ implies $u=\cos\big(\frac{\pi}{2}-\cos^{-1}v\big)=\cos(\sin^{-1}v)$, hence $u^{2}+v^{2}=1$ with $u\ge 0$.

$ (\sqrt{3}x)^{2}+x^{2}=1 \Rightarrow 4x^{2}=1 \Rightarrow x^{2}=\frac{1}{4}$

So $x=\pm\frac{1}{2}$, but the condition $u=\sqrt{3}x\ge 0$ excludes $x=-\frac{1}{2}$.

Also $x=\frac{1}{2}$ satisfies domain $|\sqrt{3}x|\le 1,\;|x|\le 1$.

Final answer: $x=\frac{1}{2}$