The number of points of intersection of the curves `2y =1 " and " y = "sin" x, -2 pi le x le 2 pi`, is

Options

A. 2
B. 3
C. 4
D. 1

Answer

To find the number of points of intersection of the curves given by the equations \(2y = 1\) and \(y = \sin x\) in the interval \(-2\pi \leq x \leq 2\pi\), we can follow these steps:

### Step 1: Simplify the first equation
The first equation is given as:
\[ 2y = 1 \]
Dividing both sides by 2, we find:
\[ y = \frac{1}{2} \]

### Step 2: Set the two equations equal
Now we have two equations:
1. \(y = \frac{1}{2}\)
2. \(y = \sin x\)

To find the points of intersection, we set these two equations equal to each other:
\[ \sin x = \frac{1}{2} \]

### Step 3: Solve for \(x\)
The sine function equals \(\frac{1}{2}\) at specific angles. In the interval from \(-2\pi\) to \(2\pi\), we can find these angles:
- The general solutions for \(\sin x = \frac{1}{2}\) are:
\[ x = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad x = \frac{5\pi}{6} + 2k\pi \]
where \(k\) is any integer.

### Step 4: Find specific solutions in the interval
Now we need to find the specific values of \(x\) that lie within the interval \(-2\pi \leq x \leq 2\pi\).

1. For \(k = -1\):
- \(x = \frac{\pi}{6} - 2\pi = \frac{\pi}{6} - \frac{12\pi}{6} = -\frac{11\pi}{6}\)
- \(x = \frac{5\pi}{6} - 2\pi = \frac{5\pi}{6} - \frac{12\pi}{6} = -\frac{7\pi}{6}\)

2. For \(k = 0\):
- \(x = \frac{\pi}{6}\)
- \(x = \frac{5\pi}{6}\)

3. For \(k = 1\):
- \(x = \frac{\pi}{6} + 2\pi = \frac{\pi}{6} + \frac{12\pi}{6} = \frac{13\pi}{6}\) (not in the interval)
- \(x = \frac{5\pi}{6} + 2\pi = \frac{5\pi}{6} + \frac{12\pi}{6} = \frac{17\pi}{6}\) (not in the interval)

### Step 5: List all valid solutions
The valid solutions in the interval \(-2\pi \leq x \leq 2\pi\) are:
1. \(x = -\frac{11\pi}{6}\)
2. \(x = -\frac{7\pi}{6}\)
3. \(x = \frac{\pi}{6}\)
4. \(x = \frac{5\pi}{6}\)

### Conclusion
Thus, the total number of points of intersection of the curves \(2y = 1\) and \(y = \sin x\) in the given interval is:
\[ \boxed{4} \]