Two real numbers $x$ and $y$ are such that $8y^4+4x^2y^2+4xy^2+2x^3+2y^2+2x=x^2+1$.
Find all possible values of $x+2y^2$
1. Start with the given equation:
   \[
   8y^4 + 4x^2y^2 + 4xy^2 + 2x^3 + 2y^2 + 2x = x^2 + 1
   \]

2. Rearrange the equation to bring all terms to one side:
   \[
   8y^4 + 4x^2y^2 + 4xy^2 + 2x^3 + 2y^2 + 2x - x^2 - 1 = 0
   \]

3. Observe that the equation can be factored. Let's try to factor it:
   \[
   (x^2 + 2y^2 + 1)(4y^2 + 2x - 1) = 0
   \]

4. Analyze the factors separately:
   - For the first factor, \(x^2 + 2y^2 + 1 = 0\):
     \[
     x^2 + 2y^2 + 1 \geq 1 > 0
     \]
     Since \(x^2\) and \(2y^2\) are non-negative, \(x^2 + 2y^2 + 1\) cannot be zero. Therefore, this factor does not contribute any solutions.

   - For the second factor, \(4y^2 + 2x - 1 = 0\):
     \[
     4y^2 + 2x - 1 = 0
     \]
     Solving for \(x\):
     \[
     2x = 1 - 4y^2
     \]
     \[
     x = \frac{1}{2} - 2y^2
     \]

5. We need to find the value of \(x + 2y^2\):
   \[
   x + 2y^2 = \left(\frac{1}{2} - 2y^2\right) + 2y^2 = \frac{1}{2}
   \]

Conclusion:
The only possible value of \(x + 2y^2\) that satisfies the given equation is \(\boxed{\frac{1}{2}}\).