How do you calculate \[{\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)\] ?

Hint: The question is involving the trigonometric function. The cosine or cos is one of the trigonometry ratios. Here in this question, given an inverse cosine function, then using the specified angle of trigonometric cosine ratios we can get the required value of angle \[\theta \] .

Complete step-by-step answer:
The trigonometry and inverse trigonometry are reversed to each other. We have six different trigonometry ratios in the trigonometry.
Cosine or cos is the one of the trigonometric function defined as the ratio between the adjacent side and hypotenuse of right angled triangle with the angle \[\theta \]
The value of specified angles of the cos function are
 \[\cos {0^ \circ } = 1\]
 \[\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}\]
 \[\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}\]
 \[\cos {60^ \circ } = \dfrac{1}{2}\]
 \[\cos {90^ \circ } = 0\]
Now, Consider the given equation
 \[ \Rightarrow \,\,\,{\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)\]
 \[\dfrac{{\sqrt 3 }}{2}\] is the value of angle \[\cos {30^ \circ }\]
 \[ \Rightarrow \,\,{\cos ^{ - 1}}\left( {\,\cos {{30}^ \circ }} \right)\]
As we know now the \[x.{x^{ - 1}} = 1\] , then
 \[ \Rightarrow \,\,{1.30^ \circ }\]
 \[\therefore \,\,{\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) = {30^ \circ }\]
Therefore, angle \[\theta = {30^ \circ }\]
Now we convert angle \[\theta \] degree to radian by multiplying \[\dfrac{\pi }{{180}}\] , then
 \[ \Rightarrow \,\,\theta = 30 \times \dfrac{\pi }{{180}}\]
 \[\therefore ,\theta = {\dfrac{\pi }{6}^c}\]
However, the cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from \[2\pi \] to find the solution in the fourth quadrant.
 \[ \Rightarrow \,\,\theta = 2\pi - \dfrac{\pi }{6}\]
By taking 6 has LCM in RHS
 \[ \Rightarrow \,\,\theta = \dfrac{{12\pi - \pi }}{6}\]
 \[\therefore ,\theta = {\dfrac{{11\pi }}{6}^c}\]
The period of the \[\cos (\theta )\] function is \[2\pi \] so value \[\dfrac{{\sqrt 3 }}{2}\] will repeat every \[2\pi \] radians in both directions.
 \[ \Rightarrow \theta = \dfrac{\pi }{6} + 2n\pi ,\,\,\,\dfrac{{11\pi }}{6} + 2n\pi \] , for any integer \[n\] .
However, since the domain of the \[{\cos ^{ - 1}}\] is [-1,1] , \[\theta = \dfrac{\pi }{6}\] is the only one solution.
So, the correct answer is “$\dfrac{\pi }{6}$”.

Note: The question is related about the inverse trigonometry. The inverse trigonometry is represented as arc, inv or the trigonometry ratio raised to the power -1. We must be familiar with the table of trigonometry ratios for the standard angle, then we can find the required solution for the given question.