To convert the given polar equation $r = 1 - 2 \cos(\theta)$ into rectangular form, we use the relationships between polar and rectangular coordinates:

$x = r \cos(\theta), \quad y = r \sin(\theta), \quad r^2 = x^2 + y^2, \quad \cos(\theta) = \frac{x}{r}.$

Step-by-Step Solution:

1. Substitute $\cos(\theta) = \frac{x}{r}$ into the polar equation: $r = 1 - 2 \cos(\theta).$ Replacing $\cos(\theta)$ gives: $r = 1 - 2 \left( \frac{x}{r} \right).$

2. Eliminate the fraction by multiplying through by $r$: $r^2 = r - 2x.$

3. Replace $r^2$ with $x^2 + y^2$: $x^2 + y^2 = r - 2x.$

4. Solve for $r$: Recall $r = \sqrt{x^2 + y^2}$, so substitute $r$ back: $x^2 + y^2 = \sqrt{x^2 + y^2} - 2x.$

5. Square both sides to eliminate the square root: $(x^2 + y^2)^2 = \left( \sqrt{x^2 + y^2} - 2x \right)^2.$

   Expand the right-hand side using $(a - b)^2 = a^2 - 2ab + b^2$: $(x^2 + y^2)^2 = (\sqrt{x^2 + y^2})^2 - 4x\sqrt{x^2 + y^2} + 4x^2.$

   Simplify further: $(x^2 + y^2)^2 = x^2 + y^2 - 4x\sqrt{x^2 + y^2} + 4x^2.$

6. Simplify and rearrange: Combine like terms: $(x^2 + y^2)^2 - x^2 - y^2 - 4x^2 = -4x\sqrt{x^2 + y^2}.$

This is the rectangular equation. You may leave the result in this implicit form since solving explicitly for $y$ as a function of $x$ is generally complicated.

Summary:

The rectangular form of the given polar equation $r = 1 - 2 \cos(\theta)$ simplifies to: $(x^2 + y^2)^2 = (x^2 + y^2) - 4x\sqrt{x^2 + y^2} + 4x^2.$

Would you like a detailed explanation of any step? Let me know if you'd like to focus on simplifying further or any clarifications!

Related Questions:

1. How do you convert $r = 1 + 2 \sin(\theta)$ to rectangular form?
2. What is the relationship between polar and rectangular coordinates?
3. How do you plot $r = 1 - 2 \cos(\theta)$ on a polar graph?
4. Can you derive $\cos(\theta)$ and $\sin(\theta)$ using $x$ and $y$?
5. How do you handle square roots when converting polar equations?

Tip: When converting between polar and rectangular forms, start with the key relationships $r^2 = x^2 + y^2$ and $\cos(\theta) = \frac{x}{r}$.