If ` r sin theta = 3, r = 4 (1 + sin theta), 0le theta le 2 pi" then " theta` =

Options

A. `pi//6, pi//3`
B. `pi//6, 5pi//6`
C. `pi/3, pi//4`
D. `pi//2, pi`

Answer

To solve the equation \( r \sin \theta = 3 \) and \( r = 4(1 + \sin \theta) \) for \( 0 \leq \theta \leq 2\pi \), we can follow these steps:

### Step 1: Substitute the value of \( r \)
We know that \( r = 4(1 + \sin \theta) \). We can substitute this into the first equation:
\[ 4(1 + \sin \theta) \sin \theta = 3 \]

### Step 2: Expand the equation
Expanding the left-hand side gives:
\[ 4 \sin \theta + 4 \sin^2 \theta = 3 \]

### Step 3: Rearrange the equation
Rearranging the equation to set it to zero:
\[ 4 \sin^2 \theta + 4 \sin \theta - 3 = 0 \]

### Step 4: Use the quadratic formula
This is a quadratic equation in terms of \( \sin \theta \). We can use the quadratic formula \( \sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = 4 \), and \( c = -3 \):
\[ \sin \theta = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 4 \cdot (-3)}}{2 \cdot 4} \]

### Step 5: Calculate the discriminant
Calculating the discriminant:
\[ \sqrt{16 + 48} = \sqrt{64} = 8 \]

### Step 6: Solve for \( \sin \theta \)
Now substituting back into the quadratic formula:
\[ \sin \theta = \frac{-4 \pm 8}{8} \]
This gives us two solutions:
1. \( \sin \theta = \frac{4}{8} = \frac{1}{2} \)
2. \( \sin \theta = \frac{-12}{8} = -\frac{3}{2} \) (not valid since sine cannot be less than -1)

### Step 7: Find \( \theta \) for valid \( \sin \theta \)
For \( \sin \theta = \frac{1}{2} \):
\[ \theta = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad \theta = \frac{5\pi}{6} + 2n\pi \]
where \( n \) is any integer.

### Step 8: Find solutions in the given range
Since we are looking for solutions in the range \( 0 \leq \theta \leq 2\pi \):
1. \( \theta = \frac{\pi}{6} \)
2. \( \theta = \frac{5\pi}{6} \)
3. \( \theta = \frac{13\pi}{6} \) (which is \( \frac{\pi}{6} + 2\pi \))
4. \( \theta = \frac{17\pi}{6} \) (which is \( \frac{5\pi}{6} + 2\pi \))

However, \( \frac{13\pi}{6} \) and \( \frac{17\pi}{6} \) exceed \( 2\pi \), so we discard them.

### Final Solutions
Thus, the valid solutions for \( \theta \) in the range \( 0 \leq \theta \leq 2\pi \) are:
\[ \theta = \frac{\pi}{6}, \frac{5\pi}{6} \]