$\sin ^{-1}(2 x)+\sin ^{-1} x= \frac{\pi}{3}$ Here are the other options randomized with the correct answer for the given MCQ question:

Solve:$\begin{array}{l}\Rightarrow \sin ^{-1}(2 x)=\frac{\pi}{3}-\sin ^{-1} x \\\Rightarrow 2 x=\sin \left(\frac{\pi}{3}-\sin ^{-1} x\right) \\\Rightarrow 2 x=\sin \frac{\pi}{3} \cos \left(\sin ^{-1} x\right)-\cos \frac{\pi}{3} \sin \left(\sin ^{-1} x\right) \\\Rightarrow 2 x=\frac{\sqrt{3}}{2} \cos \left(\cos ^{-1} \sqrt{1-x^{2}}\right)-\frac{1}{2} \cdot x \\\left.\Rightarrow 4 x=\sqrt{3} \sqrt{1-x^{2}}\right)-x \\\left.\Rightarrow 5 x=\sqrt{3} \sqrt{1-x^{2}}\right)-x \Rightarrow 25 x^{2}=3\left(1-x^{2}\right) \\\Rightarrow 28 x^{2}=3 \Rightarrow x= \pm \sqrt{\frac{3}{28}} \\\therefore x=-\sqrt{\frac{3}{28}} \text { dose not comply the equation } \\\therefore x=\sqrt{\frac{3}{28}}\end{array}$