Plz solve the following

$$\begin{gathered} {\cos }^{-1}x + {\sin }^{-1}\frac{x}{2} =\frac{\mathrm{\pi }}{6} \\ \Rightarrow {\cos }^{-1}x +\frac{\mathrm{\pi }}{2}-{\cos }^{-1}\frac{x}{2}=\frac{\mathrm{\pi }}{6} [ \mathrm{Since}, {\sin }^{-1}\frac{\mathrm{x}}{2}+{\cos }^{-1}\frac{\mathrm{x}}{2}=\frac{\mathrm{\pi }}{2}] \\ \Rightarrow {\cos }^{-1}\mathrm{x} -{\cos }^{-1}\frac{\mathrm{x}}{2}=\frac{-\mathrm{\pi }}{3} \\ \Rightarrow {\cos }^{-1}\mathrm{x} = {\cos }^{-1}\frac{\mathrm{x}}{2}-{\cos }^{-1}\frac{1}{2} \\ \Rightarrow {\cos }^{-1}\mathrm{x}={\cos }^{-1}[\frac{\mathrm{x}}{2}\times \frac{1}{2}+\frac{\sqrt{3}}{2}\sqrt{1-\frac{{\mathrm{x}}^2}{4}} ] \\ \Rightarrow \mathrm{x} = \frac{\mathrm{x}}{4}+\frac{\sqrt{3}}{4}\sqrt{4-{\mathrm{x}}^2} \\ \Rightarrow \frac{3\mathrm{x}}{4}=\frac{\sqrt{3}}{4}\sqrt{4-{\mathrm{x}}^2} \\ \Rightarrow 3{\mathrm{x}}^2=4-{\mathrm{x}}^2 \\ \Rightarrow 4{\mathrm{x}}^2=4 \\ \Rightarrow \mathrm{x} =\pm 1 \end{gathered}$$

But -1 is not satisfying the given equation , so x =1 is the only solution.