Solve 4cos22 x 3 0 over the interval 0 pi x pi12 x 11 pi

Solve 4cos^2(2 x) - 3 = 0| over the interval [0, pi)|. x = pi/12, x = 11 pi/12| x = pi/12| x = pi/12, x = 5 pi/12| x = 5 pi/12, x = 7pi/12| x = pi/12, x = 5 pi/12, x = 7 pi/12, x = 11 pi/125| None of the above.

Solution

4cos²2x-3=0

4cos²2x=3

cos²2x=3/4

cos2x=(+-sqrt3)/2

2x=cos^-1(+-(sqrt3)/2)

2x=pi/6,5pi/6,7pi,6,11pi/6

x=pi/12,5pi/12,7pi/12,11pi/12

Correct option is e