#cos^2 (x/2) = cos^2 x#
Use trig identity #cos x= 2cos^2 (x/2) - 1#.
Call #cos (x/2) = t #. We have:
#t^2 = (2t^2 - 1)^2#
#t^2 = 4t^4 - 4t^2 + 1#
#4t^4 - 5t^2 + 1 = 0#
Solve this bi-quadratic equation in t. Call t^2 = T, we get:
4T^2 - 5T + 1 = 0
Since a + b + c = 0, use Shortcut. The 2 real roots are T = 1 and
# T = c/a = 1/4#
A.#T = t^2 = 1 #--> #t = +- 1#
a. #cos (x/2) = t = 1 #--> arc #x/2 = 0# --> #x = 0#
b. #cos (x /2) = t = -1# --> #x/2 = pi #--> #x = 2pi#
B. #T = t^2 = 1/4# --> #t = +- 1/2#
a. #t = cos (x/2) = 1/2# --> arc #x/2 = +- pi/3# --> #x = +- 2pi/3# -->
x = (2pi)/3 and x = (4pi/3)
b. #t = cos (x/2) = -1/2 #--> #x/2 = +- (3pi)/4# --> x = +- (3pi)/2 -->
x = pi/2 and x = (3pi)/2
Answers:# 0, pi/2, (2pi)/3, (4pi)/3, (3pi)/2, 2pi#
