Find the number of solutions of `|cos x|=sin x , 0 le x le 4 pi`

Answer

To find the number of solutions for the equation \( | \cos x | = \sin x \) in the interval \( 0 \leq x \leq 4\pi \), we can follow these steps:

### Step 1: Understand the Functions
We need to analyze the functions \( | \cos x | \) and \( \sin x \). The function \( \sin x \) oscillates between 0 and 1, while \( | \cos x | \) oscillates between 0 and 1 as well but is always non-negative.

### Step 2: Set Up the Equation
The equation \( | \cos x | = \sin x \) can be split into two cases:
1. \( \cos x = \sin x \)
2. \( -\cos x = \sin x \)

### Step 3: Solve the First Case
For the first case, \( \cos x = \sin x \):
- This can be rewritten as \( \tan x = 1 \).
- The general solutions for this equation are given by:
\[ x = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]
- Within the interval \( 0 \leq x \leq 4\pi \), we can find the specific solutions:
- For \( n = 0 \): \( x = \frac{\pi}{4} \)
- For \( n = 1 \): \( x = \frac{5\pi}{4} \)
- For \( n = 2 \): \( x = \frac{9\pi}{4} \)
- For \( n = 3 \): \( x = \frac{13\pi}{4} \)

Thus, we have 4 solutions from this case.

### Step 4: Solve the Second Case
For the second case, \( -\cos x = \sin x \):
- This can be rewritten as \( \cos x + \sin x = 0 \).
- The general solutions for this equation are given by:
\[ x = \frac{3\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]
- Within the interval \( 0 \leq x \leq 4\pi \), we can find the specific solutions:
- For \( n = 0 \): \( x = \frac{3\pi}{4} \)
- For \( n = 1 \): \( x = \frac{7\pi}{4} \)
- For \( n = 2 \): \( x = \frac{11\pi}{4} \)
- For \( n = 3 \): \( x = \frac{15\pi}{4} \)

Thus, we have 4 solutions from this case as well.

### Step 5: Count the Total Solutions
Now, we combine the solutions from both cases:
- From \( \cos x = \sin x \): 4 solutions
- From \( -\cos x = \sin x \): 4 solutions

### Final Answer
The total number of solutions in the interval \( 0 \leq x \leq 4\pi \) is:
\[ \text{Total Solutions} = 4 + 4 = 8 \]