Find the number of solution of the equation `1+e^cot^(2x)=sqrt(2|sinx|-1)+(1-cos2x)/(1+sin^4x)forx in (0,5pi)dot`

Answer

To solve the equation \( 1 + e^{\cot^2 x} = \sqrt{2|\sin x| - 1} + \frac{1 - \cos 2x}{1 + \sin^4 x} \) for \( x \) in the interval \( (0, 5\pi) \), we will analyze both sides of the equation step by step.

### Step 1: Analyze the Left-Hand Side (LHS)
The left-hand side of the equation is given by:
\[ LHS = 1 + e^{\cot^2 x} \]
Since \( \cot^2 x \) is always non-negative, \( e^{\cot^2 x} \) will always be greater than or equal to 1. Therefore, the minimum value of \( LHS \) occurs when \( \cot^2 x = 0 \):
\[ LHS_{\text{min}} = 1 + e^0 = 1 + 1 = 2 \]
As \( \cot^2 x \) increases, \( e^{\cot^2 x} \) approaches infinity, making \( LHS \) unbounded:
\[ LHS_{\text{max}} = \infty \]

### Step 2: Analyze the Right-Hand Side (RHS)
The right-hand side of the equation is:
\[ RHS = \sqrt{2|\sin x| - 1} + \frac{1 - \cos 2x}{1 + \sin^4 x} \]

#### Part 1: Analyze \( \sqrt{2|\sin x| - 1} \)
The term \( |\sin x| \) varies between 0 and 1. Thus:
- When \( |\sin x| = 0 \), \( \sqrt{2 \cdot 0 - 1} \) is undefined (not valid).
- When \( |\sin x| = 1 \), \( \sqrt{2 \cdot 1 - 1} = \sqrt{1} = 1 \).

Therefore, the minimum value of \( \sqrt{2|\sin x| - 1} \) occurs when \( |\sin x| \) is just slightly greater than \( 0.5 \) (i.e., \( |\sin x| = 0.5 \)):
\[ \sqrt{2 \cdot 0.5 - 1} = \sqrt{0} = 0 \]
The maximum value occurs at \( |\sin x| = 1 \):
\[ \sqrt{2 \cdot 1 - 1} = 1 \]

#### Part 2: Analyze \( \frac{1 - \cos 2x}{1 + \sin^4 x} \)
Using the identity \( \cos 2x = 1 - 2\sin^2 x \):
\[ 1 - \cos 2x = 2\sin^2 x \]
Thus:
\[ RHS = \sqrt{2|\sin x| - 1} + \frac{2\sin^2 x}{1 + \sin^4 x} \]
The term \( \frac{2\sin^2 x}{1 + \sin^4 x} \) is always non-negative and reaches its maximum when \( \sin^2 x \) is maximized.

### Step 3: Find Maximum Value of RHS
- The maximum value of \( \sqrt{2|\sin x| - 1} \) is \( 1 \).
- The maximum value of \( \frac{2\sin^2 x}{1 + \sin^4 x} \) can be evaluated, but it will not exceed \( 1 \) as well.

Thus, the maximum value of \( RHS \) is:
\[ RHS_{\text{max}} = 1 + 1 = 2 \]

### Step 4: Equate LHS and RHS
From our analysis:
- \( LHS \) has a minimum value of \( 2 \) and can go to infinity.
- \( RHS \) has a maximum value of \( 2 \).

The only point where \( LHS = RHS \) is when both equal \( 2 \).

### Step 5: Solve for \( x \)
Set \( LHS = 2 \):
\[ 1 + e^{\cot^2 x} = 2 \implies e^{\cot^2 x} = 1 \implies \cot^2 x = 0 \implies \cot x = 0 \]
The solutions for \( \cot x = 0 \) occur at:
\[ x = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z} \]

### Step 6: Count Solutions in \( (0, 5\pi) \)
The values of \( x \) in the interval \( (0, 5\pi) \):
- For \( n = 0 \): \( x = \frac{\pi}{2} \)
- For \( n = 1 \): \( x = \frac{3\pi}{2} \)
- For \( n = 2 \): \( x = \frac{5\pi}{2} \)
- For \( n = 3 \): \( x = \frac{7\pi}{2} \)
- For \( n = 4 \): \( x = \frac{9\pi}{2} \)

This gives us a total of **5 solutions**.

### Final Answer
The number of solutions of the equation in the interval \( (0, 5\pi) \) is **5**.