#color(red)cos(color(blue)cos(x))=sqrt(3)/2#
Let #y=cos(x)#
#color(red)cos(color(blue)y)=sqrt(3)/2#
We remember that #color(red)cos(color(blue)(pi/6))
=cos(30^@)=sqrt3/2# - It's one of the special angles
Thus #y=pi/6# is one solution.
But the domain of trig functions extend to all real numbers, so it's useful to recall the graph of #cos# (with a line #y=sqrt3/2#).
graph{(y-cos(x))(y-sqrt3/2)=0 [-2, 6, -2, 2]}
Due to parity
#cos(-x)=cos(x)#
and periodicity
#cos(x+2pi)=cos(x)#
of this function, the solutions are
#y=+-pi/6+2kpi#
where #k in ZZ#
But #y=cos(x)#, therefore #-1<=y<=1#
Only solutions
#y_1=pi/6 or y_2=-pi/6#
are left.
First case
#cos(x)=pi/6# - the same problem as before, but it's not a special angle
#x=+-arccos(pi/6)+2k_1pi# where #k_1 in ZZ#
Second case
#cos(x)=-pi/6#
#x=+-arccos(-pi/6)+2k_2pi# where #k_2 in ZZ#
#x=+-(pi-arccos(pi/6))+2k_2pi# (#arccos# property)
#x=+-arccos(pi/6)+(2k_2+-1)pi#
#x=+-arccos(pi/6)+(2k_2+1)pi# (set of solutions unchanged)
Union of sets of solutions from both cases gives us
#x=+-arccos(pi/6)+k_3pi# where #k_3 in ZZ#
Graph of arccos (with a line #x=pi/6#)
graph{(x-cos(y))(x-pi/6)sqrt(y(pi-y))=0 [-4, 4, -0.5, 3.5]}
Useful links
https://www.wolframalpha.com/
https://www.wolframalpha.com/input/?i=arccos(pi%2F6)
