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

Answer

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

### Step 1: Understand the Functions
We need to analyze the two functions:
- \( y_1 = 2^{\cos x} \)
- \( y_2 = |\sin x| \)

### Step 2: Determine the Range of Each Function
1. **For \( y_1 = 2^{\cos x} \)**:
- The cosine function \( \cos x \) varies from -1 to 1.
- Therefore, \( 2^{\cos x} \) will vary from \( 2^{-1} = \frac{1}{2} \) to \( 2^{1} = 2 \).
- Thus, the range of \( y_1 \) is \( \left[\frac{1}{2}, 2\right] \).

2. **For \( y_2 = |\sin x| \)**:
- The sine function \( \sin x \) varies from -1 to 1, so \( |\sin x| \) varies from 0 to 1.
- Thus, the range of \( y_2 \) is \( [0, 1] \).

### Step 3: Sketch the Graphs
- **Graph of \( y_1 = 2^{\cos x} \)**:
- Starts at \( 2 \) when \( x = 0 \) (since \( \cos(0) = 1 \)).
- Decreases to \( \frac{1}{2} \) at \( x = \pi \) (since \( \cos(\pi) = -1 \)).
- Increases back to \( 2 \) at \( x = 2\pi \).

- **Graph of \( y_2 = |\sin x| \)**:
- Starts at \( 0 \) at \( x = 0 \), reaches \( 1 \) at \( x = \frac{\pi}{2} \), returns to \( 0 \) at \( x = \pi \), reaches \( 1 \) again at \( x = \frac{3\pi}{2} \), and returns to \( 0 \) at \( x = 2\pi \).

### Step 4: Analyze Intersections
- The graphs of \( y_1 \) and \( y_2 \) will intersect where \( 2^{\cos x} = |\sin x| \).
- Since \( y_1 \) ranges from \( \frac{1}{2} \) to \( 2 \) and \( y_2 \) ranges from \( 0 \) to \( 1 \), we focus on the interval where both functions can intersect, which is \( [\frac{1}{2}, 1] \).

### Step 5: Count the Solutions
- In the interval \( [-2\pi, 2\pi] \), the sine function completes two full cycles.
- Each cycle of \( |\sin x| \) intersects \( 2^{\cos x} \) at four points:
- Two intersections where \( |\sin x| \) is increasing to 1 and decreasing back to 0.
- Two intersections where \( |\sin x| \) is decreasing from 1 back to 0 and increasing again.

### Conclusion
- Therefore, the total number of solutions in the interval \( [-2\pi, 2\pi] \) is \( 4 \) (from the first cycle) + \( 4 \) (from the second cycle) = **8 solutions**.