Find the number of solution of the equations
`2^(cos x)=|sin x|, ` when ` x in[-2pi,2pi]`

Answer

To solve the equation \( 2^{\cos x} = |\sin x| \) for \( x \) in the interval \([-2\pi, 2\pi]\), we can follow these steps:

### Step 1: Analyze the functions
We need to analyze the two sides of the equation:
- The left side, \( 2^{\cos x} \), varies between \( 2^{-1} = \frac{1}{2} \) (when \( \cos x = -1 \)) and \( 2^{1} = 2 \) (when \( \cos x = 1 \)).
- The right side, \( |\sin x| \), varies between \( 0 \) and \( 1 \).

### Step 2: Determine the range of solutions
Since \( 2^{\cos x} \) ranges from \( \frac{1}{2} \) to \( 2 \) and \( |\sin x| \) ranges from \( 0 \) to \( 1 \), we can conclude that the equation \( 2^{\cos x} = |\sin x| \) can only have solutions when \( |\sin x| \) is in the interval \([\frac{1}{2}, 1]\).

### Step 3: Find critical points of \( |\sin x| \)
The points where \( |\sin x| = 1 \) occur at:
- \( x = \frac{\pi}{2} + 2n\pi \)
- \( x = -\frac{\pi}{2} + 2n\pi \)

The points where \( |\sin x| = \frac{1}{2} \) occur at:
- \( x = \frac{\pi}{6} + n\pi \)
- \( x = \frac{5\pi}{6} + n\pi \)

### Step 4: Identify solutions in the interval \([-2\pi, 2\pi]\)
Now we will find all the solutions in the interval \([-2\pi, 2\pi]\):

1. For \( |\sin x| = 1 \):
- \( x = \frac{\pi}{2} \) (in the interval)
- \( x = -\frac{\pi}{2} \) (in the interval)

2. For \( |\sin x| = \frac{1}{2} \):
- \( x = \frac{\pi}{6} \) (in the interval)
- \( x = \frac{5\pi}{6} \) (in the interval)
- \( x = -\frac{5\pi}{6} \) (in the interval)
- \( x = -\frac{\pi}{6} \) (in the interval)

### Step 5: Count the solutions
Now we list the solutions we found:
- \( x = \frac{\pi}{2} \)
- \( x = -\frac{\pi}{2} \)
- \( x = \frac{\pi}{6} \)
- \( x = \frac{5\pi}{6} \)
- \( x = -\frac{5\pi}{6} \)
- \( x = -\frac{\pi}{6} \)

This gives us a total of **6 solutions** in the interval \([-2\pi, 2\pi]\).

### Final Answer
The number of solutions of the equation \( 2^{\cos x} = |\sin x| \) in the interval \([-2\pi, 2\pi]\) is **6**.