To find the x-coordinate of the point \((x, y)\) where \(y = 2\sqrt{2}\) on the unit circle, we start from the equation of the unit circle, which is given by:
\[ x^2 + y^2 = 1 \]
Substituting \(y = 2\sqrt{2}\) into the equation:
\[ x^2 + (2\sqrt{2})^2 = 1 \]
Calculating \((2\sqrt{2})^2\):
\[ (2\sqrt{2})^2 = 4 \cdot 2 = 8 \]
Now substitute this value back into the equation:
\[ x^2 + 8 = 1 \]
To isolate \(x^2\):
\[ x^2 = 1 - 8 \] \[ x^2 = -7 \]
Since \(x^2 = -7\) does not yield a real value for \(x\), it indicates that the point \((x, y) = (x, 2\sqrt{2})\) cannot lie on the unit circle.
Therefore, there is no x-coordinate for the given y-coordinate of \(2\sqrt{2}\) on the unit circle. The coordinates do not intersect the unit circle.
Final Answer: No valid x-coordinate exists (y = 2√2 does not lie on the unit circle).