General solution of the equation
`2 cot ^(2) theta + 2sqrt3 cot theta + 4 cosec theta + 8 =0` is

Options

A. `theta = n pi pm (pi)/(6) , n in I`
B. ` n pi + (pi)/(6) , n in I`
C. ` 2 n pi + (pi)/( 6) , n In I`
D. `2 n pi + (11 pi)/(6), n in I`

Answer

To solve the equation \(2 \cot^2 \theta + 2\sqrt{3} \cot \theta + 4 \csc \theta + 8 = 0\), we will follow these steps:

### Step 1: Simplify the Equation
We can start by dividing the entire equation by 2 to simplify it:

\[ \cot^2 \theta + \sqrt{3} \cot \theta + 2 \csc \theta + 4 = 0 \]

### Step 2: Use Trigonometric Identities
Recall the identity \(\cot^2 \theta + 1 = \csc^2 \theta\). We can express \(\cot^2 \theta\) in terms of \(\csc^2 \theta\):

\[ \cot^2 \theta = \csc^2 \theta - 1 \]

Substituting this into the equation gives:

\[ (\csc^2 \theta - 1) + \sqrt{3} \cot \theta + 2 \csc \theta + 4 = 0 \]

### Step 3: Rearranging the Equation
This simplifies to:

\[ \csc^2 \theta + \sqrt{3} \cot \theta + 2 \csc \theta + 3 = 0 \]

### Step 4: Introduce Substitutions
Let \(x = \cot \theta\) and \(y = \csc \theta\). Then, we can express \(\csc^2 \theta\) as \(y^2\) and \(\cot \theta\) as \(x\):

\[ y^2 + \sqrt{3}x + 2y + 3 = 0 \]

### Step 5: Solve for \(y\)
Now, we can rearrange this to solve for \(y\):

\[ y^2 + 2y + 3 + \sqrt{3}x = 0 \]

This is a quadratic equation in \(y\). We can use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Here, \(a = 1\), \(b = 2\), and \(c = 3 + \sqrt{3}x\):

\[ y = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (3 + \sqrt{3}x)}}{2 \cdot 1} \]

### Step 6: Calculate the Discriminant
The discriminant is:

\[ 4 - 4(3 + \sqrt{3}x) = 4 - 12 - 4\sqrt{3}x = -8 - 4\sqrt{3}x \]

For \(y\) to be real, the discriminant must be non-negative:

\[ -8 - 4\sqrt{3}x \geq 0 \implies \sqrt{3}x \leq -2 \implies x \leq -\frac{2}{\sqrt{3}} \]

### Step 7: Find General Solution
From the conditions \(x = \cot \theta = -\sqrt{3}\) and \(y = \csc \theta = -2\):

1. \(\cot \theta = -\sqrt{3} \implies \theta = \frac{5\pi}{6} + n\pi\)
2. \(\csc \theta = -2 \implies \theta = \frac{7\pi}{6} + 2n\pi\) or \(\theta = \frac{11\pi}{6} + 2n\pi\)

### Final General Solution
Thus, the general solution for the equation is:

\[ \theta = n\pi - \frac{\pi}{6} \quad \text{or} \quad \theta = n\pi + \frac{\pi}{6}, \quad n \in \mathbb{Z} \]