The formula states that #cos(2x)=cos^2(x)-sin^2(x)#. So, the equation becomes
#cos^2(x)-sin^2(x)-cos(x)=0#.
We can use the fundamental identity #cos^2(x)+sin^2(x)=1# to obtain #sin^2(x)=1-cos^2(x)# and write the equation in terms of #cos(x)# only:
#cos^2(x)-(1-cos^2(x))-cos(x)=0#
#iff#
#cos^2(x)-1+cos^2(x)-cos(x)=0#
#iff#
#2cos^2(x)-cos(x)-1=0#
Now we use a variable #t=cos(x)# to solve the quadratic equation:
#2cos^2(x)-cos(x)-1=0 -> 2t^2-t-1=0#
The solution of the equation in #t# are #t=1# and #t=-1/2#.
So, we must find values of #x# for which #cos(x)=1# and #cos(x)=-1/2#.
The first is easy, since #cos(x)=1# if and only if #x=2kpi#, for some #k \in \mathbb{Z}#.
As for #cos(x)=-1/2#, we have that the fundamental solutions are #\pm{2pi}/3#, and each one repeats every #2pi#.
