The image contains two sets of problems: one asks to convert rectangular equations to polar form, and the other asks to convert polar equations to rectangular form. Here are the solutions for each problem:

2.1 Rectangular to Polar Conversion

The basic relations between rectangular coordinates $(x, y)$ and polar coordinates $(r, \theta)$ are: $x = r \cos \theta, \quad y = r \sin \theta, \quad r = \sqrt{x^2 + y^2}, \quad \text{and} \quad \tan \theta = \frac{y}{x}$

2.1.1 $y = 3x + 2$

Convert to polar form using $x = r \cos \theta$ and $y = r \sin \theta$: $r \sin \theta = 3r \cos \theta + 2$ Dividing by $r$ (assuming $r \neq 0$): $\sin \theta = 3 \cos \theta + \frac{2}{r}$ Or equivalently: $r = \frac{2}{\sin \theta - 3 \cos \theta}$

2.1.2 $y^2 = 3 - x^2$

Convert to polar form using $x^2 = r^2 \cos^2 \theta$ and $y^2 = r^2 \sin^2 \theta$: $r^2 \sin^2 \theta = 3 - r^2 \cos^2 \theta$ Rearranging: $r^2 (\sin^2 \theta + \cos^2 \theta) = 3$ Since $\sin^2 \theta + \cos^2 \theta = 1$: $r^2 = 3 \quad \text{so} \quad r = \sqrt{3}$

2.1.3 $x^2 + y^2 = 6y$

Convert to polar form using $x^2 + y^2 = r^2$ and $y = r \sin \theta$: $r^2 = 6r \sin \theta$ Divide by $r$ (assuming $r \neq 0$): $r = 6 \sin \theta$

2.2 Polar to Rectangular Conversion

The basic relations for this conversion are the same as mentioned above.

2.2.1 $r = 2 \cos \theta$

Convert to rectangular form using $r = \sqrt{x^2 + y^2}$ and $\cos \theta = \frac{x}{r}$: $\sqrt{x^2 + y^2} = 2 \cdot \frac{x}{\sqrt{x^2 + y^2}}$ Square both sides: $x^2 + y^2 = 4x$ Rearrange: $x^2 - 4x + y^2 = 0$

2.2.2 $r = 2 \cos 2\theta$

Use the double-angle identity $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$ and convert to rectangular form: $r = 2(\cos^2 \theta - \sin^2 \theta)$ Substitute $\cos \theta = \frac{x}{r}$ and $\sin \theta = \frac{y}{r}$: $\sqrt{x^2 + y^2} = 2\left(\frac{x^2 - y^2}{x^2 + y^2}\right)$ Multiply by $\sqrt{x^2 + y^2}$: $(x^2 + y^2)^2 = 2(x^2 - y^2)$

2.2.3 $r = \frac{3}{1 - 2\cos \theta}$

Convert to rectangular form by multiplying by $1 - 2\cos \theta$ and then using $\cos \theta = \frac{x}{r}$: $r(1 - 2 \cos \theta) = 3$ Substitute $r = \sqrt{x^2 + y^2}$ and $\cos \theta = \frac{x}{r}$: $\sqrt{x^2 + y^2} - 2x = 3$ Square both sides: $(x^2 + y^2) - 4x\sqrt{x^2 + y^2} + 4x^2 = 9$

Do you want any further details or explanations on any part? Here are some related questions you might find helpful:

1. How do you derive the relation between polar and rectangular coordinates?
2. What are some common challenges when converting between polar and rectangular forms?
3. Can all equations be converted between these forms?
4. How do trigonometric identities help in conversions?
5. What are some practical applications of using polar coordinates?

Tip: When converting between polar and rectangular forms, always start by writing down the fundamental relationships between $x$, $y$, $r$, and $\theta$. This will guide your conversion process smoothly.