How do you solve #cos x - 3 cos (x/2) = 0#?

#cosx-3cos(x/2)=0#

#cos(x/2times2)-3cos(x/2)=0#

Double angle formula #cos2theta=cos^2theta-sin^2theta#
#cos^2(x/2)-sin^2(x/2)-3cos(x/2)=0#

Rearrange so that #cos^2theta-sin^2theta=cos^2theta-(1-cos^2theta)=cos^2theta-1+cos^2theta=2cos^2theta-1#
#2cos^2(x/2)-3cos(x/2)-1=0#

Using quadratic formula:

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#cos(x/2)=(3+-sqrt(9-4(2)(-1)))/4#

#cos(x/2)=(3+-sqrt17)/4#

#x/2=cos^(-1)((3+-sqrt17)/4)#

#x/2=cos^(-1)((3-sqrt17)/4)# only

Why? Well, #(3+sqrt17)/4# is greater than 1 and since the domain of #cos^(-1)x# is #-1< x<1#, then there will be no solution

Now, if the domain was between #0 < x<360# then it would become #0< x/2<180#

#x/2=106'18'#

#x=212'37'#

#cos x - 3cos (x/2) = 0#
Note: #cos x = 2co^2 (x/2) - 1# (trig identity)
#2cos^2 (x/2) - 3cos (x/2) - 1 = 0#.
Solve this quadratic equation for #cos (x/2)#.
#D = d^2 = b^2 - 4ac = 9 + 8 = 17# --> #d = +- sqrt17#
There are 2 real roots:
#cos (x/2) = -b/(2a) +- d/(2a) = 3/4 +- sqrt17/4#
#cos (x/2) = (3 + sqrt17)/4# (rejected because > 1)
#cos (x/2) = (3 - sqrt17)/4 = - 0.28#
Calculator and unit circle gives 2 solutions for #x/2#.
#x/2 = +- 106^@30#

a. #x/2 = 106^@30 + k360^@#
#x = 212^@60 + k720^@#
b. #x/2 = - 106^@30#, or its co-terminal
#x/2 = 360 - 106.30 = 253^@70 + k360^@#
#x = 507.4 = 507.4 - 360 = 147^@4 + k720^@#