Solve for x, \[{{\cos }^{-1}}x+{{\sin }^{-1}}\left( \dfrac{x}{2} \right)=\dfrac{\pi }{6}\]

Hint: In this question, we first need to convert the inverse cosine term into inverse sine term using the properties of inverse trigonometric functions. Then use the formula of difference of two inverse sine functions from the properties of inverse trigonometric functions to simplify it further. Now, from the standard values of sine functions we get the inverse value.
\[{{\cos }^{-1}}x+{{\sin }^{-1}}x=\dfrac{\pi }{2}\]
\[{{\sin }^{-1}}x-{{\sin }^{-1}}y={{\sin }^{-1}}\left\{ x\sqrt{1-{{y}^{2}}}-y\sqrt{1-{{x}^{2}}} \right\}\]

Complete step-by-step answer:
Now, from the given expression in the question we have
\[\Rightarrow {{\cos }^{-1}}x+{{\sin }^{-1}}\left( \dfrac{x}{2} \right)=\dfrac{\pi }{6}\]
Let us now consider the left hand side of the above expression
\[\Rightarrow {{\cos }^{-1}}x+{{\sin }^{-1}}\left( \dfrac{x}{2} \right)\]
As we already know from the properties of the inverse trigonometric functions that inverse cosine function can be written in terms of inverse sine function as follows
\[{{\cos }^{-1}}x=\dfrac{\pi }{2}-{{\sin }^{-1}}x\]
Now, using this property the above expression can be further written as
\[\Rightarrow \dfrac{\pi }{2}-{{\sin }^{-1}}x+{{\sin }^{-1}}\left( \dfrac{x}{2} \right)\]
Now, this expression can also be written as
\[\Rightarrow \dfrac{\pi }{2}+{{\sin }^{-1}}\left( \dfrac{x}{2} \right)-{{\sin }^{-1}}x\]
The formula for sum of two inverse sine functions is given by
\[{{\sin }^{-1}}x-{{\sin }^{-1}}y={{\sin }^{-1}}\left\{ x\sqrt{1-{{y}^{2}}}-y\sqrt{1-{{x}^{2}}} \right\}\]
Now, on comparing the above expression with this formula we get
\[x=\dfrac{x}{2},y=x\]
Now, on substituting these values back in the formula we get,
\[\Rightarrow \dfrac{\pi }{2}+{{\sin }^{-1}}\left\{ \dfrac{x}{2}\sqrt{1-{{x}^{2}}}-x\sqrt{1-{{\left( \dfrac{x}{2} \right)}^{2}}} \right\}\]
Now, this can be further written as
\[\Rightarrow \dfrac{\pi }{2}+{{\sin }^{-1}}\left\{ \dfrac{x}{2}\sqrt{1-{{x}^{2}}}-\dfrac{x}{2}\sqrt{4-{{x}^{2}}} \right\}\]
Now, on equating this to the right hand side we have
\[\Rightarrow \dfrac{\pi }{2}+{{\sin }^{-1}}\left\{ \dfrac{x}{2}\sqrt{1-{{x}^{2}}}-\dfrac{x}{2}\sqrt{4-{{x}^{2}}} \right\}=\dfrac{\pi }{6}\]
Now, on rearranging the terms in the above equation we get,
\[\Rightarrow {{\sin }^{-1}}\left\{ \dfrac{x}{2}\sqrt{1-{{x}^{2}}}-\dfrac{x}{2}\sqrt{4-{{x}^{2}}} \right\}=-\dfrac{\pi }{3}\]
As we already know from the properties of trigonometric and inverse trigonometric functions that
\[\begin{align}
  & \sin \left( {{\sin }^{-1}}\theta \right)=\theta \\
 & \sin \left( -\theta \right)=-\sin \theta \\
\end{align}\]
Now, on applying the sine function on both the sides we get,
\[\Rightarrow \sin \left( {{\sin }^{-1}}\left\{ \dfrac{x}{2}\sqrt{1-{{x}^{2}}}-\dfrac{x}{2}\sqrt{4-{{x}^{2}}} \right\} \right)=\sin \left( -\dfrac{\pi }{3} \right)\]
Now, from the above mentioned properties we can rewrite it as
\[\Rightarrow \dfrac{x}{2}\sqrt{1-{{x}^{2}}}-\dfrac{x}{2}\sqrt{4-{{x}^{2}}}=-\dfrac{\sqrt{3}}{2}\text{ }\left[ \because \sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2} \right]\]
Now, this can be further written as
\[\Rightarrow x\sqrt{1-{{x}^{2}}}-x\sqrt{4-{{x}^{2}}}=-\sqrt{3}\]
Now, on rearranging the terms we get,
\[\Rightarrow x\sqrt{1-{{x}^{2}}}+\sqrt{3}=x\sqrt{4-{{x}^{2}}}\]
Let us now do the squaring on both sides
\[\Rightarrow {{x}^{2}}\left( 1-{{x}^{2}} \right)+3+2\sqrt{3}\times x\sqrt{1-{{x}^{2}}}={{x}^{2}}\left( 4-{{x}^{2}} \right)\text{ }\]
Now, on rearranging the terms on both sides we get,
\[\Rightarrow 2\sqrt{3}\times x\sqrt{1-{{x}^{2}}}={{x}^{2}}\left( 4-{{x}^{2}}-1+{{x}^{2}} \right)-3\]
Now, on further simplification we get,
\[\Rightarrow 2\sqrt{3}\times x\sqrt{1-{{x}^{2}}}=-3\left( 1-{{x}^{2}} \right)\]
Now, on cancelling out the common terms on both the sides we get,
\[\Rightarrow 2x=-\sqrt{3}\sqrt{1-{{x}^{2}}}\]
Now, by squaring on both sides we get,
\[\Rightarrow 4{{x}^{2}}=3\left( 1-{{x}^{2}} \right)\]
Now, on rearranging the terms and simplifying we get,
\[\Rightarrow 7{{x}^{2}}=3\]
Let us now divide with 7 on both sides and apply square root
\[\begin{align}
  & \Rightarrow \sqrt{{{x}^{2}}}=\sqrt{\dfrac{3}{7}} \\
 & \therefore x=\pm \sqrt{\dfrac{3}{7}} \\
\end{align}\]

Note:Instead of converting the inverse cosine function we can take it to the other side and write the constant value in terms of inverse cosine function. Then by using the sum of two inverse cosine functions we can solve it further. Both the methods give the same result.
It is important to note that while rearranging the terms and applying the properties we should not neglect any of the terms or write the incorrect value because it changes the result accordingly.