The equation given is in polar form:

$r = 2 + \sin(0)$

First, let's simplify the equation. Since $\sin(0) = 0$, the equation becomes:

$r = 2$

This represents a circle with radius $r = 2$ centered at the origin in polar coordinates.

In Cartesian coordinates, the point $r = 2$ with $\theta = 0$ corresponds to the point $(x, y) = (2, 0)$.

Now, the question asks how many times this curve passes through $x = 1$ in Cartesian coordinates.

For a circle with radius $2$, the equation in Cartesian coordinates is:

$x^2 + y^2 = r^2 = 4$

We want to know how many times this circle intersects the vertical line $x = 1$.

Substitute $x = 1$ into the equation of the circle:

$1^2 + y^2 = 4$ $1 + y^2 = 4$ $y^2 = 3$ $y = \pm \sqrt{3}$

Thus, the circle intersects the line $x = 1$ at two points: $(1, \sqrt{3})$ and $(1, -\sqrt{3})$.

So, the curve $r = 2 + \sin(0)$ passes through $x = 1$ twice in Cartesian coordinates.

Would you like more details or have any questions?

Here are 5 related questions:

1. How would the curve change if $r = 2 + \sin(\theta)$?
2. What is the equation of a general circle in polar coordinates?
3. How do you convert between polar and Cartesian coordinates?
4. What is the significance of the sine term in polar equations?
5. How can you find the area enclosed by a polar curve?

Tip: To convert polar coordinates to Cartesian coordinates, use $x = r \cos(\theta)$ and $y = r \sin(\theta)$.