How do you solve cos^2 (x/2) = cos^2 x, within the interval [0,2pi)?

1 Answer
Nghi N.
Feb 17, 2016

0, pi/2, (2pi)/3, (4pi)/3, (3pi)/2, and 2pi

Explanation:

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

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