To find the value of \( x \) for the point \( (x, y) \) where \( y = 2\sqrt{2} \) and the point lies on the unit circle, we start with the equation of the unit circle:
\[ x^2 + y^2 = 1 \]
Given \( y = 2\sqrt{2} \), we can substitute this value into the unit circle equation:
\[ x^2 + (2\sqrt{2})^2 = 1 \]
Calculating \( (2\sqrt{2})^2 \):
\[ (2\sqrt{2})^2 = 4 \cdot 2 = 8 \]
Now substitute this back into the equation:
\[ x^2 + 8 = 1 \]
Next, we isolate \( x^2 \):
\[ x^2 = 1 - 8 \] \[ x^2 = -7 \]
Since \( x^2 = -7 \) does not yield a real number (as \( x^2 \) cannot be negative in the real number system), this indicates that the value of \( y = 2\sqrt{2} \) does not correspond to a point on the unit circle in Quadrant I. The maximum value for \( y \) on the unit circle is \( 1 \), thus \( 2\sqrt{2} \) exceeds that limit.
Therefore, there is no valid value for \( x \) corresponding to \( y = 2\sqrt{2} \) on the unit circle.