Solve : `1+2cosec x=-("sec"^(2)(x)/(2))/(2)`

Explanation

To solve the equation \( 1 + 2 \csc x = -\frac{\sec^2 x}{2} \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 1 + 2 \csc x = -\frac{\sec^2 x}{2} \]
We can express \(\csc x\) and \(\sec^2 x\) in terms of sine and cosine:
\[ 1 + \frac{2}{\sin x} = -\frac{1}{2 \cos^2 x} \]

### Step 2: Clear the fractions
To eliminate the fractions, we can multiply through by \(2 \sin x \cos^2 x\):
\[ 2 \sin x \cos^2 x (1 + \frac{2}{\sin x}) = 2 \sin x \cos^2 x \left(-\frac{1}{2 \cos^2 x}\right) \]
This simplifies to:
\[ 2 \sin x \cos^2 x + 4 \cos^2 x = -\sin x \]

### Step 3: Rearranging the equation
Rearranging gives us:
\[ 2 \sin x \cos^2 x + \sin x + 4 \cos^2 x = 0 \]
Factoring out \(\sin x\):
\[ \sin x (2 \cos^2 x + 1) + 4 \cos^2 x = 0 \]

### Step 4: Set each factor to zero
We can set each factor to zero:
1. \(\sin x = 0\)
2. \(2 \cos^2 x + 1 + 4 \cos^2 x = 0\)

### Step 5: Solve \(\sin x = 0\)
The solutions for \(\sin x = 0\) are:
\[ x = n\pi, \quad n \in \mathbb{Z} \]

### Step 6: Solve \(2 \cos^2 x + 4 \cos^2 x + 1 = 0\)
This simplifies to:
\[ 6 \cos^2 x + 1 = 0 \]
\[ \cos^2 x = -\frac{1}{6} \]
Since \(\cos^2 x\) cannot be negative, there are no solutions from this factor.

### Final Solution
Thus, the only solutions to the original equation are:
\[ x = n\pi, \quad n \in \mathbb{Z} \]