cos^2 (x/2) = cos^2 xcos2(x2)=cos2x
Use trig identity cos x= 2cos^2 (x/2) - 1cosx=2cos2(x2)−1.
Call cos (x/2) = t cos(x2)=t. We have:
t^2 = (2t^2 - 1)^2t2=(2t2−1)2
t^2 = 4t^4 - 4t^2 + 1t2=4t4−4t2+1
4t^4 - 5t^2 + 1 = 04t4−5t2+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 = 1/4T=ca=14
A.T = t^2 = 1 T=t2=1--> t = +- 1t=±1
a. cos (x/2) = t = 1 cos(x2)=t=1--> arc x/2 = 0x2=0 --> x = 0x=0
b. cos (x /2) = t = -1cos(x2)=t=−1 --> x/2 = pi x2=π--> x = 2pix=2π
B. T = t^2 = 1/4T=t2=14 --> t = +- 1/2t=±12
a. t = cos (x/2) = 1/2t=cos(x2)=12 --> arc x/2 = +- pi/3x2=±π3 --> x = +- 2pi/3x=±2π3 -->
x = (2pi)/3 and x = (4pi/3)
b. t = cos (x/2) = -1/2 t=cos(x2)=−12--> x/2 = +- (3pi)/4x2=±3π4 --> x = +- (3pi)/2 -->
x = pi/2 and x = (3pi)/2
Answers: 0, pi/2, (2pi)/3, (4pi)/3, (3pi)/2, 2pi0,π2,2π3,4π3,3π2,2π
