`{x inR" :"|cosx|gesinx}nn[0,(3pi)/(2)]=`

Options

A. `[0,(pi)/(4)]uu[(3pi)/(4),(3pi)/(2)]`
B. `[0,(pi)/(4)]uu[(pi)/(2),(3pi)/(2)]`
C. `[0,pi/4]uu[(5pi)/(4),(3pi)/(2)]`
D. `[0,(3pi)/(2)]`

Answer

To solve the problem `{x in R : |cos x| ≥ sin x} ∩ [0, (3π)/2]`, we will follow these steps:

### Step 1: Understand the inequality
We need to find the values of \( x \) such that \( |cos x| \geq sin x \). This means we need to analyze the graphs of \( |cos x| \) and \( sin x \).

### Step 2: Analyze the graphs
1. **Graph of \( |cos x| \)**:
- The function \( cos x \) oscillates between -1 and 1.
- The graph of \( |cos x| \) will reflect the negative part of \( cos x \) above the x-axis, thus it will always be non-negative.

2. **Graph of \( sin x \)**:
- The function \( sin x \) starts at 0, reaches 1 at \( \frac{\pi}{2} \), and goes back to 0 at \( \pi \), then goes negative until \( \frac{3\pi}{2} \).

### Step 3: Find points of intersection
To find where \( |cos x| = sin x \):
- Set \( cos x = sin x \) and solve for \( x \):
\[ \tan x = 1 \implies x = \frac{\pi}{4} + n\pi \]
In the interval \( [0, \frac{3\pi}{2}] \), this gives us \( x = \frac{\pi}{4} \).

- Set \( -cos x = sin x \) (since \( |cos x| \) can be negative):
\[ -cos x = sin x \implies cos x = -sin x \implies \tan x = -1 \implies x = \frac{3\pi}{4} + n\pi \]
In the interval \( [0, \frac{3\pi}{2}] \), this gives us \( x = \frac{3\pi}{4} \).

### Step 4: Determine intervals
Now we need to determine the intervals where \( |cos x| \geq sin x \):
1. From \( 0 \) to \( \frac{\pi}{4} \), \( |cos x| \) is greater than \( sin x \).
2. From \( \frac{\pi}{4} \) to \( \frac{3\pi}{4} \), \( |cos x| \) is less than \( sin x \).
3. From \( \frac{3\pi}{4} \) to \( \frac{3\pi}{2} \), \( |cos x| \) is greater than \( sin x \).

### Step 5: Combine intervals
The solution set where \( |cos x| \geq sin x \) in the interval \( [0, \frac{3\pi}{2}] \) is:
\[ [0, \frac{\pi}{4}] \cup [\frac{3\pi}{4}, \frac{3\pi}{2}] \]

### Final Answer
Thus, the final answer is:
\[ \{ x \in [0, \frac{3\pi}{2}] : |cos x| \geq sin x \} = [0, \frac{\pi}{4}] \cup [\frac{3\pi}{4}, \frac{3\pi}{2}] \]