First notice that we can deduct some things from the equation:

#cos(cos^-1(-pi/3))#

We can't have #cos^-1(-pi/3)# , it is outside the domain of the function #f(x)=cos^-1(x)#.

Howeve, because #cos^-1(cos(x))=x# , the only way I see to move on, is to apply the same concept.

That means the equation from above tells us, that at some point, some angle was converted into #-pi/3# radiens.

Imagine #-pi/3# radiens on the unit circle. This angle is not within #0<=theta<=pi# . In fact this translates to the upper half of the unit circle (see illustrations).

You can see that:

#Cos(-pi/3)=cos(2pi-pi/3)#

Now calculate #cos(2pi-pi/3)=0.5#

Using your prefered graph tool, you can also see that #cos(x)=0.5# has many more solutions. One particular solution exist in between #cos(-pi/3)# and #cos(2pi-pi/3)#, that is #cos(pi/3)#.

This means:

#cos(-pi/3)=cos(2pi-pi/3)=cos(pi/3)#

As you can see:

only one solution satisfies, #0<=theta<=pi# , that is #theta = pi/3#

Imagine #theta=pi/3= cos^-1(cos(pi/3))=cos^-1(cos(-pi/3))# , because #cos(pi/3)=cos(-pi/3)#

and so if #cos(cos^-1(u))=u# then we can write:

#u=-pi/3#

#cos^-1(cos(-pi/3))=cos^-1(cos(u))=u=cos(cos^-1(u))=cos(cos^-1(-pi/3))#

To conclude, when #theta=pi/3# , it satisfies #0<=theta<=pi# , and that can be rewritten to form the equation #cos(cos^-1(-pi/3))=pi/3# if #cos(cos^-1(u))=u#

Idon't know... It makes sense for me to do it this way, but I would like someone to double check it for fundamental flaws :)