The most general value of `theta` which satisfies both the equation in General solution of `(1 + 2sin theta)^2 + (sqrt (3)tantheta-1)^2` = `0` is

Options

A. `n pi + pi/6`
B. `2 n pi + (11pi)/6`
C. `2n pi + (7pi)/6`
D. none

Answer

To solve the equation \((1 + 2\sin \theta)^2 + (\sqrt{3}\tan \theta - 1)^2 = 0\), we will follow these steps:

### Step 1: Set the equation to zero
Since the sum of two squares is zero only when both squares are individually zero, we can set each term to zero:
1. \((1 + 2\sin \theta)^2 = 0\)
2. \((\sqrt{3}\tan \theta - 1)^2 = 0\)

### Step 2: Solve the first equation
From the first equation:
\[ 1 + 2\sin \theta = 0 \]
Solving for \(\sin \theta\):
\[ 2\sin \theta = -1 \implies \sin \theta = -\frac{1}{2} \]

### Step 3: Find general solutions for \(\sin \theta = -\frac{1}{2}\)
The sine function is negative in the third and fourth quadrants. The angles where \(\sin \theta = -\frac{1}{2}\) are:
\[ \theta = \frac{7\pi}{6} + 2n\pi \quad \text{and} \quad \theta = \frac{11\pi}{6} + 2n\pi \quad (n \in \mathbb{Z}) \]

### Step 4: Solve the second equation
From the second equation:
\[ \sqrt{3}\tan \theta - 1 = 0 \]
Solving for \(\tan \theta\):
\[ \sqrt{3}\tan \theta = 1 \implies \tan \theta = \frac{1}{\sqrt{3}} \]

### Step 5: Find general solutions for \(\tan \theta = \frac{1}{\sqrt{3}}\)
The tangent function is positive in the first and third quadrants. The angles where \(\tan \theta = \frac{1}{\sqrt{3}}\) are:
\[ \theta = \frac{\pi}{6} + n\pi \quad (n \in \mathbb{Z}) \]

### Step 6: Combine the solutions
Now we have two sets of solutions:
1. From \(\sin \theta = -\frac{1}{2}\):
- \(\theta = \frac{7\pi}{6} + 2n\pi\)
- \(\theta = \frac{11\pi}{6} + 2n\pi\)

2. From \(\tan \theta = \frac{1}{\sqrt{3}}\):
- \(\theta = \frac{\pi}{6} + n\pi\)

### Step 7: General solution
The most general solution for \(\theta\) that satisfies both equations can be expressed as:
\[ \theta = 2n\pi + \frac{7\pi}{6} \quad \text{or} \quad \theta = 2n\pi + \frac{11\pi}{6} \quad \text{or} \quad \theta = n\pi + \frac{\pi}{6} \quad (n \in \mathbb{Z}) \]

### Final Answer
The most general value of \(\theta\) which satisfies both equations is:
\[ \theta = 2n\pi + \frac{7\pi}{6} \quad \text{or} \quad \theta = 2n\pi + \frac{11\pi}{6} \quad \text{or} \quad \theta = n\pi + \frac{\pi}{6} \quad (n \in \mathbb{Z}) \]