By definition of a function arccos(x), its value is defined as such that, if used as an argument to a function cos(.), the result will be x.
In other words, by definition of arccos(x),
cos(arccos(x))=x
Of course, this is valid only for those x, that can theoretically be the value of cos(.), that is -1 <= x <= 1, which is a domain of a function arccos(x).
Because of periodicity of function cos(.), the values of arccos(x) are restricted to an interval [0,pi] to assure proper one-to-one correspondence between argument and a function's value.
Thus, for example, since
cos(pi/4)=sqrt(2)/2
then, by definition of arccos,
arccos(sqrt(2)/2)=pi/4
Let's use this definition for our problem by applying a function cos(.) to both sides of a given equation.
cos(arccos(x-pi/3))=cos(pi/6)
The left side, by definition of arccos(.), equals to x-pi/3, while the right side equals to sqrt(3)/2.
Therefore,
x-pi/3=sqrt(3)/2
Solution for x is
x=pi/3+sqrt(3)/2
Complete guide to properties of function arccos(x) and other inverse trigonometric functions can be found at Unizor following the menu items "Trigonometry" - "Inverse Trigonometric Functions".
