Solve `rsinθ=3` , `r=4(1+sinθ) `, `0≤θ≤2π.`

Answer

To solve the equations \( r \sin \theta = 3 \) and \( r = 4(1 + \sin \theta) \) for \( 0 \leq \theta \leq 2\pi \), we will eliminate \( r \) and find the values of \( \theta \).

### Step 1: Substitute \( r \) from the second equation into the first equation.
We know that:
\[ r = 4(1 + \sin \theta) \]
Substituting this into the first equation gives:
\[ 4(1 + \sin \theta) \sin \theta = 3 \]

### Step 2: Expand the equation.
Expanding the left side, we have:
\[ 4 \sin \theta + 4 \sin^2 \theta = 3 \]

### Step 3: Rearrange the equation.
Rearranging the equation gives us:
\[ 4 \sin^2 \theta + 4 \sin \theta - 3 = 0 \]

### Step 4: Factor the quadratic equation.
Now we will factor the quadratic equation:
\[ (2 \sin \theta + 3)(2 \sin \theta - 1) = 0 \]

### Step 5: Solve for \( \sin \theta \).
Setting each factor to zero gives us:
1. \( 2 \sin \theta + 3 = 0 \) → \( \sin \theta = -\frac{3}{2} \) (not possible since sine values range from -1 to 1)
2. \( 2 \sin \theta - 1 = 0 \) → \( \sin \theta = \frac{1}{2} \)

### Step 6: Find the values of \( \theta \).
The solutions for \( \sin \theta = \frac{1}{2} \) in the interval \( 0 \leq \theta \leq 2\pi \) are:
\[ \theta = \frac{\pi}{6}, \quad \theta = \frac{5\pi}{6} \]

### Final Answer:
Thus, the solutions for \( \theta \) are:
\[ \theta = \frac{\pi}{6}, \quad \theta = \frac{5\pi}{6} \]