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".
