Find all values of `theta ` lying between `0` and `2 pi` satisfying the equation `r sin theta = sqrt 3` and `r + 4 sin theta =2(sqrt 3+1)`

Answer

To solve the equations \( r \sin \theta = \sqrt{3} \) and \( r + 4 \sin \theta = 2(\sqrt{3} + 1) \), we can follow these steps:

### Step 1: Express \( r \) in terms of \( \sin \theta \)
From the first equation, we can express \( r \) as:
\[ r = \frac{\sqrt{3}}{\sin \theta} \]

### Step 2: Substitute \( r \) into the second equation
Now, substitute \( r \) into the second equation:
\[ \frac{\sqrt{3}}{\sin \theta} + 4 \sin \theta = 2(\sqrt{3} + 1) \]

### Step 3: Clear the fraction by multiplying through by \( \sin \theta \)
Multiplying both sides by \( \sin \theta \) gives:
\[ \sqrt{3} + 4 \sin^2 \theta = 2(\sqrt{3} + 1) \sin \theta \]

### Step 4: Rearrange the equation
Rearranging the equation results in:
\[ 4 \sin^2 \theta - 2(\sqrt{3} + 1) \sin \theta + \sqrt{3} = 0 \]

### Step 5: Identify coefficients for the quadratic formula
This is a quadratic equation in terms of \( \sin \theta \):
\[ a = 4, \quad b = -2(\sqrt{3} + 1), \quad c = \sqrt{3} \]

### Step 6: Apply the quadratic formula
Using the quadratic formula \( \sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ \sin \theta = \frac{2(\sqrt{3} + 1) \pm \sqrt{(-2(\sqrt{3} + 1))^2 - 4 \cdot 4 \cdot \sqrt{3}}}{2 \cdot 4} \]

### Step 7: Simplify the expression
Calculating the discriminant:
\[ b^2 = 4(\sqrt{3} + 1)^2 = 4(3 + 2\sqrt{3} + 1) = 4(4 + 2\sqrt{3}) = 16 + 8\sqrt{3} \]
\[ 4ac = 16\sqrt{3} \]
Thus,
\[ b^2 - 4ac = (16 + 8\sqrt{3}) - 16\sqrt{3} = 16 - 8\sqrt{3} \]

Now substituting back:
\[ \sin \theta = \frac{2(\sqrt{3} + 1) \pm \sqrt{16 - 8\sqrt{3}}}{8} \]

### Step 8: Calculate the two possible values for \( \sin \theta \)
1. **First value**:
\[ \sin \theta = \frac{2(\sqrt{3} + 1) + \sqrt{16 - 8\sqrt{3}}}{8} \]
2. **Second value**:
\[ \sin \theta = \frac{2(\sqrt{3} + 1) - \sqrt{16 - 8\sqrt{3}}}{8} \]

### Step 9: Find the angles \( \theta \)
Now we need to find \( \theta \) for both values of \( \sin \theta \) within the interval \( [0, 2\pi] \).

1. For \( \sin \theta = \frac{\sqrt{3}}{2} \):
- \( \theta = \frac{\pi}{3}, \frac{2\pi}{3} \)

2. For \( \sin \theta = \frac{1}{2} \):
- \( \theta = \frac{\pi}{6}, \frac{5\pi}{6} \)

### Final Values of \( \theta \)
Thus, the values of \( \theta \) that satisfy the equations are:
\[ \theta = \frac{\pi}{6}, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{5\pi}{6} \]