Question

The number of real roots of the equation ${\sin }^{-1}\left(6x\right)+{\sin }^{-1}\left(6\sqrt{3}x\right)+\frac{\pi }{2}=0$ is

Answer 1

$x\in [\frac{-1}{6\sqrt{3}},\frac{1}{6\sqrt{3}}]$

Now, Let $f\left(x\right)={\sin }^{-1}\left(6x\right)+{\sin }^{-1}\left(6\sqrt{3}x\right)+\frac{\pi }{2}$ is continuous and strictly increasing in $[\frac{-1}{6\sqrt{3}},\frac{1}{6\sqrt{3}})$.
Also, $f\left(\frac{-1}{6\sqrt{3}}\right)<0$ and $f\left(\frac{1}{6\sqrt{3}}\right)>0$
Using MVT, there is exactly one root of $f\left(x\right)=0$ in $[\frac{-1}{6\sqrt{3}},\frac{1}{6\sqrt{3}})$.