Total number of solution of `2^(cosx)=abs(sinx)"in"[ -2pi,5pi]` is equal to :

Options

A. 12
B. 14
C. 16
D. 15

Answer

To solve the equation \(2^{\cos x} = |\sin x|\) in the interval \([-2\pi, 5\pi]\), we will analyze the graphs of both sides of the equation and count the number of intersections.

### Step-by-Step Solution:

1. **Understanding the Functions**:
- The function \(2^{\cos x}\) oscillates between \(2\) (when \(\cos x = 1\)) and \(\frac{1}{2}\) (when \(\cos x = -1\)).
- The function \(|\sin x|\) oscillates between \(0\) and \(1\).

2. **Finding the Periodicity**:
- Both \(2^{\cos x}\) and \(|\sin x|\) are periodic functions with a period of \(2\pi\).
- Therefore, we can analyze the interval from \(0\) to \(2\pi\) and then extend our findings to the entire interval \([-2\pi, 5\pi]\).

3. **Graphing the Functions**:
- For \(x \in [0, 2\pi]\):
- At \(x = 0\), \(2^{\cos 0} = 2^1 = 2\) and \(|\sin 0| = 0\).
- At \(x = \frac{\pi}{2}\), \(2^{\cos \frac{\pi}{2}} = 2^0 = 1\) and \(|\sin \frac{\pi}{2}| = 1\).
- At \(x = \pi\), \(2^{\cos \pi} = 2^{-1} = \frac{1}{2}\) and \(|\sin \pi| = 0\).
- At \(x = \frac{3\pi}{2}\), \(2^{\cos \frac{3\pi}{2}} = 2^0 = 1\) and \(|\sin \frac{3\pi}{2}| = 1\).
- At \(x = 2\pi\), \(2^{\cos 2\pi} = 2^1 = 2\) and \(|\sin 2\pi| = 0\).

4. **Counting Intersections in \([0, 2\pi]\)**:
- The function \(2^{\cos x}\) starts at \(2\), decreases to \(1\) at \(\frac{\pi}{2}\), decreases to \(\frac{1}{2}\) at \(\pi\), increases to \(1\) at \(\frac{3\pi}{2}\), and returns to \(2\) at \(2\pi\).
- The function \(|\sin x|\) starts at \(0\), increases to \(1\) at \(\frac{\pi}{2}\), decreases to \(0\) at \(\pi\), increases to \(1\) at \(\frac{3\pi}{2}\), and decreases to \(0\) at \(2\pi\).
- The intersections occur at:
- Between \(0\) and \(\frac{\pi}{2}\) (1 intersection),
- At \(\frac{\pi}{2}\) (1 intersection),
- Between \(\pi\) and \(\frac{3\pi}{2}\) (1 intersection),
- At \(\frac{3\pi}{2}\) (1 intersection).
- Total intersections in \([0, 2\pi]\) = 4.

5. **Extending to the Interval \([-2\pi, 5\pi]\)**:
- The interval \([-2\pi, 0]\) will have the same number of intersections as \([0, 2\pi]\), which is 4.
- The interval \([2\pi, 4\pi]\) will also have 4 intersections.
- The interval \([4\pi, 5\pi]\) is equivalent to \([0, \pi]\), which has 2 intersections.

6. **Calculating Total Solutions**:
- Total solutions = Intersections from \([-2\pi, 0]\) + Intersections from \([0, 2\pi]\) + Intersections from \([2\pi, 4\pi]\) + Intersections from \([4\pi, 5\pi]\)
- Total solutions = \(4 + 4 + 4 + 2 = 14\).

### Final Answer:
The total number of solutions of \(2^{\cos x} = |\sin x|\) in the interval \([-2\pi, 5\pi]\) is **14**.