Solutions of `sin^(-1) (sinx) = sinx` are if `x in (0, 2pi)`

Options

A. 4 real roots
B. 2 positive real roots
C. 2 negative real roots
D. 5 real roots

Answer

To solve the equation \( \sin^{-1}(\sin x) = \sin x \) for \( x \) in the interval \( (0, 2\pi) \), we will analyze the properties of the functions involved.

### Step 1: Understand the function \( \sin^{-1}(\sin x) \)

The function \( \sin^{-1}(y) \) is defined for \( y \) in the range \([-1, 1]\) and returns values in the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\). Therefore, \( \sin^{-1}(\sin x) \) will return values based on the value of \( x \).

### Step 2: Determine the intervals for \( x \)

The sine function is periodic with a period of \( 2\pi \). In the interval \( (0, 2\pi) \), the sine function will take on all values from \( 0 \) to \( 1 \) and back to \( 0 \) in the first half, and then from \( 0 \) to \(-1\) and back to \( 0 \) in the second half.

### Step 3: Analyze the behavior of \( \sin^{-1}(\sin x) \)

- For \( x \in (0, \frac{\pi}{2}) \), \( \sin^{-1}(\sin x) = x \).
- For \( x \in (\frac{\pi}{2}, \frac{3\pi}{2}) \), \( \sin^{-1}(\sin x) = \pi - x \).
- For \( x \in (\frac{3\pi}{2}, 2\pi) \), \( \sin^{-1}(\sin x) = x - 2\pi \).

### Step 4: Set up the equations based on intervals

Now we can set up the equations based on the intervals:

1. In the interval \( (0, \frac{\pi}{2}) \):
\[ x = \sin x \]

2. In the interval \( (\frac{\pi}{2}, \frac{3\pi}{2}) \):
\[ \pi - x = \sin x \]

3. In the interval \( (\frac{3\pi}{2}, 2\pi) \):
\[ x - 2\pi = \sin x \]

### Step 5: Solve each equation

1. **For \( x = \sin x \)** in \( (0, \frac{\pi}{2}) \):
- This equation has one solution at \( x = 0 \) (not included in the interval) and one solution at \( x = \frac{\pi}{2} \) (included).

2. **For \( \pi - x = \sin x \)** in \( (\frac{\pi}{2}, \frac{3\pi}{2}) \):
- This equation can be rewritten as \( x = \pi - \sin x \). This will have two solutions: one in \( (\frac{\pi}{2}, \pi) \) and one in \( (\pi, \frac{3\pi}{2}) \).

3. **For \( x - 2\pi = \sin x \)** in \( (\frac{3\pi}{2}, 2\pi) \):
- This equation can be rewritten as \( x = 2\pi + \sin x \). This will not yield any solutions since \( \sin x \) is always less than or equal to 1, and \( x \) is greater than \( 2\pi \).

### Step 6: Count the solutions

From the analysis:
- We have one solution from the first interval.
- We have two solutions from the second interval.
- No solutions from the third interval.

Thus, the total number of solutions in the interval \( (0, 2\pi) \) is \( 1 + 2 = 3 \).

### Conclusion

The solutions of \( \sin^{-1}(\sin x) = \sin x \) in the interval \( (0, 2\pi) \) are three real roots.