To simplify the expression sin^2(π+x) + cos^2(2π-x), we can utilize trigonometric identities.

Recall the Pythagorean identity: sin^2(x) + cos^2(x) = 1.

Applying this identity to sin^2(π+x) and cos^2(2π-x), we get:

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

Now, substitute these expressions back into the original equation:

1 - cos^2(π+x) + 1 - sin^2(2π-x) = 0
1 - cos^2(π+x) + 1 - sin^2(-x) = 0
1 - cos^2(π+x) + 1 - sin^2(x) = 0
1 - cos^2(π+x) - sin^2(x) + 1 = 0

Now, using the Pythagorean identity again: cos^2(x) = 1 - sin^2(x), we can simplify further:

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

Therefore, the given expression sin^2(π+x) + cos^2(2π-x) is not equal to 0.