The general value of x satisfying the equation
` 2 cot^(2) x + 2 sqrt(3) cot x + 4 "cosec" x + 8 = 0 ` is : (where `n in I` )

Options

A. ` n pi - (pi)/(6) `
B. ` n pi + (pi)/(6) `
C. ` 2n pi - (pi)/(6) `
D. ` 2n pi + (pi)/(6) `

Answer

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

### Step 1: Rewrite the equation using trigonometric identities
We know that \( \csc x = \frac{1}{\sin x} \) and \( \cot x = \frac{\cos x}{\sin x} \). We can rewrite \( \csc x \) in terms of \( \cot x \):
\[ \csc x = \sqrt{1 + \cot^2 x} \]
However, for simplicity, we will keep \( \csc x \) as it is for now.

### Step 2: Rearranging the equation
The equation can be rearranged:
\[ 2 \cot^2 x + 2 \sqrt{3} \cot x + 4 \csc x + 8 = 0 \]

### Step 3: Substitute \( \cot x = y \)
Let \( y = \cot x \). Then the equation becomes:
\[ 2y^2 + 2\sqrt{3}y + 4\csc x + 8 = 0 \]
This is a quadratic equation in terms of \( y \).

### Step 4: Solve for \( y \)
To solve for \( y \), we need to express \( \csc x \) in terms of \( y \). We can use the identity \( \csc^2 x = 1 + \cot^2 x \) to express \( \csc x \):
\[ \csc x = \sqrt{1 + y^2} \]
Substituting this back into the equation gives:
\[ 2y^2 + 2\sqrt{3}y + 4\sqrt{1 + y^2} + 8 = 0 \]

### Step 5: Isolate the square root
Rearranging gives:
\[ 4\sqrt{1 + y^2} = -2y^2 - 2\sqrt{3}y - 8 \]
Squaring both sides to eliminate the square root:
\[ 16(1 + y^2) = (2y^2 + 2\sqrt{3}y + 8)^2 \]

### Step 6: Expand and simplify
Expanding the right-hand side:
\[ 16 + 16y^2 = 4y^4 + 8\sqrt{3}y^3 + (4\cdot8 + 12)y^2 + 64 \]
Combine like terms and set the equation to zero.

### Step 7: Solve the resulting polynomial
This will yield a polynomial equation in \( y \). Solve for \( y \) using the quadratic formula or factoring if possible.

### Step 8: Find \( x \) from \( y \)
Once you have the values of \( y \), use \( y = \cot x \) to find \( x \):
\[ x = \cot^{-1}(y) \]

### Step 9: General solution
The general solution for \( x \) will be:
\[ x = n\pi + \cot^{-1}(y) \]
where \( n \) is any integer.

### Final Answer
After solving the polynomial and finding \( y \), we can conclude the general values of \( x \) satisfying the original equation.