Let's first expand the right side of the equation using the trigonometric identity sin^2(x) + cos^2(x) = 1:

(sin 2x + cos 2x)^2 = sin^2(2x) + 2sin(2x)cos(2x) + cos^2(2x)
(sin 2x + cos 2x)^2 = sin^2(2x) + cos^2(2x) + 2sin(2x)cos(2x) + sin^2(2x)
(sin 2x + cos 2x)^2 = 1 + 2sin(2x)cos(2x) + 1
(sin 2x + cos 2x)^2 = 2 + 2sin(2x)cos(2x)

Now, we have Sin 2x = 2 + 2sin(2x)cos(2x)

To solve for sin 2x in terms of sin x and cos x, we need to use the double angle identities:

sin 2x = 2sin(x)cos(x)

Therefore, the equation becomes:

2sin(x)cos(x) = 2 + 2sin(x)cos(x)

Subtracting 2sin(x)cos(x) from both sides:

0 = 2

This means that the original equation Sin 2x = (sin 2x + cos 2x)^2 is not valid.