The point $P$ is on the unit circle. If the $y$-coordinate of $P$ is $\frac{2}{3}$, and $P$ is in quadrant II, then
\[
x=
\]
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Final Answer: The x-coordinate of point P is \(\boxed{-0.745}\).
Step 1 ：The point P is on the unit circle, which means it satisfies the equation \(x^2 + y^2 = 1\).
Step 2 ：We know the y-coordinate of P is \(\frac{2}{3}\), and we need to find the x-coordinate.
Step 3 ：Since P is in quadrant II, the x-coordinate should be negative.
Step 4 ：We can substitute \(y = \frac{2}{3}\) into the equation of the unit circle to solve for x.
Step 5 ：After substituting and solving, we find that \(x = -0.7453559924999299\).
Step 6 ：Rounding to three decimal places, we get \(x = -0.745\).
Step 7 ：Final Answer: The x-coordinate of point P is \(\boxed{-0.745}\).
