If `cos^(-1)|sinx|gesin^(-1)|sinx|`, then the number of integral values of x in the interval `x in [0, 3pi]` are

Options

A. 7
B. 6
C. 4
D. 5

Answer

To solve the inequality \( \cos^{-1} | \sin x | \geq \sin^{-1} | \sin x | \), we will follow these steps:

### Step 1: Understand the functions involved
We know that:
- The range of \( \cos^{-1} y \) is \( [0, \pi] \) for \( y \in [0, 1] \).
- The range of \( \sin^{-1} y \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) for \( y \in [-1, 1] \).

Since \( | \sin x | \) is always between 0 and 1, both functions are defined and we can analyze the inequality.

**Hint:** Recall the ranges of the inverse trigonometric functions to understand where they are defined.

### Step 2: Analyze the equality condition
The equality \( \cos^{-1} | \sin x | = \sin^{-1} | \sin x | \) holds at specific points. We can find these points by setting \( y = | \sin x | \):
- \( \cos^{-1} y = \sin^{-1} y \) implies \( y = \frac{1}{\sqrt{2}} \) (or \( y = -\frac{1}{\sqrt{2}} \) which is irrelevant since we are dealing with absolute values).

Thus, we have:
\[ | \sin x | = \frac{1}{\sqrt{2}} \]

**Hint:** Find the angles where the sine function equals \( \frac{1}{\sqrt{2}} \).

### Step 3: Solve for \( x \)
The solutions for \( | \sin x | = \frac{1}{\sqrt{2}} \) occur at:
\[ x = n\pi + \frac{\pi}{4} \quad \text{and} \quad x = n\pi - \frac{\pi}{4} \]
for \( n \in \mathbb{Z} \).

### Step 4: Identify the intervals
We need to find \( x \) in the interval \( [0, 3\pi] \):
1. For \( n = 0 \):
- \( x = \frac{\pi}{4} \)
- \( x = -\frac{\pi}{4} \) (not in the interval)

2. For \( n = 1 \):
- \( x = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \)
- \( x = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \)

3. For \( n = 2 \):
- \( x = 2\pi + \frac{\pi}{4} = \frac{9\pi}{4} \) (not in the interval)
- \( x = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4} \)

4. For \( n = 3 \):
- \( x = 3\pi + \frac{\pi}{4} \) (not in the interval)
- \( x = 3\pi - \frac{\pi}{4} = \frac{11\pi}{4} \) (not in the interval)

### Step 5: Count the integral values
The integral values of \( x \) in the interval \( [0, 3\pi] \) that we found are:
- \( 0 \)
- \( 1 \)
- \( 2 \)
- \( 3 \)
- \( 4 \)
- \( 5 \)
- \( 6 \)
- \( 7 \)
- \( 8 \)
- \( 9 \)

Thus, the integral values of \( x \) that satisfy the original inequality are \( 0, 3, 6, 7, 9 \).

### Final Answer
The number of integral values of \( x \) in the interval \( [0, 3\pi] \) is **5**.

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