How do you solve the identity #cos 2x + cos 4x = 0#?

Apply the trig identity:# cos 4x = 2cos^2 2x - 1.#
#f(x) = cos 2x + 2cos^2 2x - 1 = 0#
Call cos 2x = t, we have to solve the quadratic equation:

2t^2 + t - 1 = 0

Since (a - b + c) = 0, the shortcut gives: #t = - 1# and #t = -c/a = 1/2#

a. #cos 2x = t = -1 #--> #2x = +- pi# --> #x = +- pi/2#
b. #cos 2x = t = 1/2# --> #2x = +- pi/3# --> #x = +- pi/6#
Within interval #(-pi, pi),# there are 6 answers:
#+- pi/2 ; +- pi/3 and +- pi/6#
Check by calculator:
#x = pi/2# -> cos 2x = cos pi = -1 ; cos 4x = cos 2pi = 1 -->
--> f(x) = -1 + 1 = 0. OK
#x = pi/6# --> #cos 2x = cos pi/3 = 1/2# ; #cos 4x = cos (2pi)/3 = -1/2#
--># f(x) = 1/2 - 1/2 = 0.# OK
#x = pi/3# --> cos 2x = cos (2pi)/3 = 1/2 ; cos 4x = cos (4pi)/3 = - 1/2
f(x) = 1/2 - 1/2 = 0. OK

Let #theta = 2x#

#cos(2x)+cos(4x)=0#
#color(white)("XXXX")#becomes
#cos(theta)+cos(2theta) = 0#

By double angle formula
#color(white)("XXXX")cos(2theta) = 2cos^2(theta)-1#

So
#cos(theta)+(2cos^2(theta) -1) = 0#

Rearranging
#2cos^2(theta)+cos(theta)-1 = 0#

Factoring
#(2cos(theta)-1)(cos(theta)+1) = 0#

#rArrcolor(red)(cos(theta) = 1/2)color(white)("XXXXXXXXXXXX")orcolor(white)("XXX")color(blue)(cos(theta)=-1)#

if restricted to #theta in [0,2pi]#
#color(red)(theta = pi/3 or theta=(5pi)/3)color(white)("XXXXXXXXXXX")orcolor(white)("XXXX")color(blue)(theta=pi)#

or in general (with #n in ZZ#)
#color(red)(theta = pi/3+n2pi or theta=(5pi)/3+n2pi)color(white)("XXX")orcolor(white)("XXXX")color(blue)(theta=pi+2pi)#

Since #theta = 2x#
#x= pi/6+npi or x = (5pi)/6+npicolor(white)("XXXXX")orcolor(white)("XXXX")x =pi/2+npi#

Apply trig identity:
#cos a + cos b = 2cos ((a + b)/2)cos ((a - b)/2)#
In this case:
#cos 2x + cos 4x = 2cos 3x.cos x = 0#
Either cos x or cos 3x should be zero.

a. cos x = 0 --> Unit circle gives 2 solutions:
#x = pi/2 + 2kpi#, and #x = (3pi)/2 + 2kpi#
b/ cos 3x = 0 --> unit circle gives 2 solutions:
#3x = pi/2 + 2kpi# --> #x = pi/6 + (2kpi)/3#
#3x = (3pi/2) + 2kpi# --> #x = pi/2 + (2kpi)/3#