2sin^2 2x=2

in the domain [0, 2pi)

Please type this as 2(sin 2x)^2=2      2sin^2 2x=2 is confusing.

Reducing this expression:  (sin 2x)^2 = 1

Taking the sqrt of both sides, sin 2x = plus or minus 1

If sin 2x = 1, then 2x must be pi/2 or 3pi/2, and so x must be pi/4 or 3pi/4.

Now you do this for sin 2x = -1.

Ci think but i'm not entirely sure so..