Consider the domain $D$ expressed by the following system inequality:
\[x^2+(y-1)^2\leq 1,\ x\geq \frac{\sqrt{2}}{3}.\]

Suppose that the line $l$ passes through the origin and the common part of $l$ and $D$ is line segment.
Find the maximum length of the line segment $L$, then when $L$ is maximized, find the value of $\cos \theta$ for the angle $\theta\ \left(0<\theta <\frac{\pi}{2}\right)$ between the $x$-axis and $l$.
1. **Identify the domain \( D \)**:
   The domain \( D \) is defined by the system of inequalities:
   \[
   x^2 + (y-1)^2 \leq 1 \quad \text{and} \quad x \geq \frac{\sqrt{2}}{3}.
   \]
   The first inequality represents a circle centered at \((0, 1)\) with radius 1. The second inequality represents a vertical line at \( x = \frac{\sqrt{2}}{3} \).

2. **Equation of the line \( l \)**:
   The line \( l \) passes through the origin and can be expressed as \( y = mx \), where \( m \) is the slope of the line.

3. **Intersection points**:
   The line \( y = mx \) intersects the circle \( x^2 + (y-1)^2 = 1 \). Substituting \( y = mx \) into the circle's equation:
   \[
   x^2 + (mx - 1)^2 = 1.
   \]
   Expanding and simplifying:
   \[
   x^2 + m^2x^2 - 2mx + 1 = 1,
   \]
   \[
   (1 + m^2)x^2 - 2mx = 0,
   \]
   \[
   x((1 + m^2)x - 2m) = 0.
   \]
   This gives two solutions for \( x \):
   \[
   x = 0 \quad \text{or} \quad x = \frac{2m}{1 + m^2}.
   \]
   For \( x = 0 \), \( y = 0 \), which is the origin. For \( x = \frac{2m}{1 + m^2} \), substituting back into \( y = mx \):
   \[
   y = m \cdot \frac{2m}{1 + m^2} = \frac{2m^2}{1 + m^2}.
   \]
   So, the intersection point is \( \left( \frac{2m}{1 + m^2}, \frac{2m^2}{1 + m^2} \right) \).

4. **Intersection with the vertical line**:
   The line \( y = mx \) intersects the vertical line \( x = \frac{\sqrt{2}}{3} \) at:
   \[
   y = m \cdot \frac{\sqrt{2}}{3}.
   \]
   So, the intersection point is \( \left( \frac{\sqrt{2}}{3}, m \cdot \frac{\sqrt{2}}{3} \right) \).

5. **Distance between the points**:
   The distance \( L \) between the points \( \left( \frac{\sqrt{2}}{3}, m \cdot \frac{\sqrt{2}}{3} \right) \) and \( \left( \frac{2m}{1 + m^2}, \frac{2m^2}{1 + m^2} \right) \) is given by:
   \[
   L^2 = \left( \frac{2m}{1 + m^2} - \frac{\sqrt{2}}{3} \right)^2 + \left( \frac{2m^2}{1 + m^2} - m \cdot \frac{\sqrt{2}}{3} \right)^2.
   \]

6. **Simplify the distance expression**:
   \[
   L^2 = \left( \frac{2m(3) - \sqrt{2}(1 + m^2)}{3(1 + m^2)} \right)^2 + \left( \frac{2m^2(3) - m\sqrt{2}(1 + m^2)}{3(1 + m^2)} \right)^2,
   \]
   \[
   L^2 = \left( \frac{6m - \sqrt{2} - \sqrt{2}m^2}{3(1 + m^2)} \right)^2 + \left( \frac{6m^2 - m\sqrt{2} - m\sqrt{2}m^2}{3(1 + m^2)} \right)^2,
   \]
   \[
   L^2 = \left( \frac{6m - \sqrt{2}(1 + m^2)}{3(1 + m^2)} \right)^2 + \left( \frac{6m^2 - m\sqrt{2}(1 + m^2)}{3(1 + m^2)} \right)^2.
   \]

7. **Maximize the distance**:
   To maximize \( L \), we need to find the critical points by setting the derivative \( f'(m) \) to zero:
   \[
   f'(m) = \frac{4(m^2 - 3\sqrt{2}m + 1)(m^3 + m - 3\sqrt{2})}{9(m^2 + 1)^2} = 0.
   \]
   Solving this, we get \( m = \sqrt{2} \).

8. **Calculate the maximum length**:
   Substituting \( m = \sqrt{2} \) back into the distance formula:
   \[
   L = \sqrt{\frac{2}{3}}.
   \]

9. **Find \( \cos \theta \)**:
   Since \( \tan \theta = m = \sqrt{2} \), we have:
   \[
   \cos \theta = \frac{1}{\sqrt{1 + \tan^2 \theta}} = \frac{1}{\sqrt{1 + 2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}.
   \]

The final answer is \( \boxed{ \cos \theta = \frac{\sqrt{3}}{3} } \).