If `rsintheta=3, r=4(1+sintheta)` where `0<=theta<=2pi` then `theta` equal to

Options

A. 0
B. 2
C. 4
D. none of these

Answer

To solve the given problem, we need to find the values of \(\theta\) given the equations:

1. \( r \sin \theta = 3 \)
2. \( r = 4(1 + \sin \theta) \)

where \(0 \leq \theta \leq 2\pi\).

### Step 1: Express \(r\) in terms of \(\sin \theta\)

From the first equation, we can express \(r\) as:
\[ r = \frac{3}{\sin \theta} \]

### Step 2: Substitute \(r\) into the second equation

Now, substitute this expression for \(r\) into the second equation:
\[ \frac{3}{\sin \theta} = 4(1 + \sin \theta) \]

### Step 3: Cross-multiply to eliminate the fraction

Cross-multiplying gives us:
\[ 3 = 4(1 + \sin \theta) \sin \theta \]
This simplifies to:
\[ 3 = 4\sin \theta + 4\sin^2 \theta \]

### Step 4: Rearrange the equation

Rearranging this equation yields:
\[ 4\sin^2 \theta + 4\sin \theta - 3 = 0 \]

### Step 5: Solve the quadratic equation

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 \(c = -3\).

Calculating the discriminant:
\[ b^2 - 4ac = 4^2 - 4 \cdot 4 \cdot (-3) = 16 + 48 = 64 \]

Now substituting into the quadratic formula:
\[ \sin \theta = \frac{-4 \pm \sqrt{64}}{2 \cdot 4} = \frac{-4 \pm 8}{8} \]

### Step 6: Find the solutions for \(\sin \theta\)

Calculating the two possible values:
1. \(\sin \theta = \frac{-4 + 8}{8} = \frac{4}{8} = \frac{1}{2}\)
2. \(\sin \theta = \frac{-4 - 8}{8} = \frac{-12}{8} = -\frac{3}{2}\)

Since \(\sin \theta\) must be in the range \([-1, 1]\), we discard \(-\frac{3}{2}\) as it is not a valid solution.

### Step 7: Determine \(\theta\) from \(\sin \theta = \frac{1}{2}\)

The solutions for \(\sin \theta = \frac{1}{2}\) within the interval \(0 \leq \theta \leq 2\pi\) are:
\[ \theta = \frac{\pi}{6} \quad \text{and} \quad \theta = \frac{5\pi}{6} \]

### Final Answer

Thus, the values of \(\theta\) are:
\[ \theta = \frac{\pi}{6}, \frac{5\pi}{6} \]