To find the value of \( x \) for the point \( (x, y) \) on the unit circle where \( y = \frac{\sqrt{2}}{2} \), we start with the equation of the unit circle:
\[ x^2 + y^2 = 1 \]
Given that \( y = \frac{\sqrt{2}}{2} \), we substitute \( y \) into the equation of the circle:
\[ x^2 + \left(\frac{\sqrt{2}}{2}\right)^2 = 1 \]
Calculating \( \left(\frac{\sqrt{2}}{2}\right)^2 \):
\[ \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} = \frac{1}{2} \]
Now we can substitute this back into the circle's equation:
\[ x^2 + \frac{1}{2} = 1 \]
To isolate \( x^2 \), subtract \( \frac{1}{2} \) from both sides:
\[ x^2 = 1 - \frac{1}{2} = \frac{1}{2} \]
Now we take the square root of both sides to find \( x \):
\[ x = \sqrt{\frac{1}{2}} \quad \text{or} \quad x = -\sqrt{\frac{1}{2}} \]
Since we are in Quadrant 1, where both \( x \) and \( y \) are positive, we have:
\[ x = \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2} \]
So the value of \( x \) is:
\[ \boxed{\frac{\sqrt{2}}{2}} \]
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