How do I find the remaining solutions for the trigonometric equation?

[Speedking96]

Homework Statement

4sin²(2x) -1 = 0 Solve over 0 --> 2pi

The attempt at a solution

(2sin(2x) + 1) (2sin(2x) - 1) = 0

2sin(2x) +1 = 0
sin(2x) = -1/2
2x = -pi/6
x = -pi/12 and -11pi/12 which = 23pi/12 and 13pi/12.

2sin(2x) - 1 = 0
sin(2x) = 1/2
2x = pi/6
x= pi/12 and 11pi/12

I have found these four solutions, however, I do not know how to get the other four solutions.

 

[Chestermiller]

4sin²(2x) -1 = 0

2(2sin²(2x)-1)+1=0

-2cos(4x)+1=0

cos(4x) = 1/2

 

[LCKurtz]

Homework Statement

4sin²(2x) -1 = 0 Solve over 0 --> 2pi

The attempt at a solution

(2sin(2x) + 1) (2sin(2x) - 1) = 0

2sin(2x) +1 = 0
sin(2x) = -1/2
2x = -pi/6
x = -pi/12 and -11pi/12 which = 23pi/12 and 13pi/12.

2sin(2x) - 1 = 0
sin(2x) = 1/2
2x = pi/6
x= pi/12 and 11pi/12

I have found these four solutions, however, I do not know how to get the other four solutions.

You have $2x = \frac \pi 6 + 2n\pi$ but also $2x = \frac {5\pi} 6 + 2n\pi$ so $x=\frac \pi {12} + n\pi$ and $x=\frac {5\pi} {12} + n\pi$. Similarly for the other one.