Let `A={theta:2"cos"^(2)theta+"sin"thetale2}and`
`B={theta:pi//2lethetale3pi//2}`. Then find the value of `A cap B`

Answer

To solve the problem, we need to find the intersection of the sets \( A \) and \( B \).

### Step 1: Define the sets
- Set \( A \) is defined as:
\[ A = \{ \theta : 2 \cos^2 \theta + \sin \theta \leq 2 \} \]
- Set \( B \) is defined as:
\[ B = \{ \theta : \frac{\pi}{2} \leq \theta \leq \frac{3\pi}{2} \} \]

### Step 2: Solve the inequality for set \( A \)
We start with the inequality:
\[ 2 \cos^2 \theta + \sin \theta \leq 2 \]
Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \), we can rewrite the inequality:
\[ 2(1 - \sin^2 \theta) + \sin \theta \leq 2 \]
This simplifies to:
\[ 2 - 2 \sin^2 \theta + \sin \theta \leq 2 \]
Subtracting 2 from both sides gives:
\[ -2 \sin^2 \theta + \sin \theta \leq 0 \]
Factoring out \( -\sin \theta \):
\[ -\sin \theta (2 \sin \theta - 1) \leq 0 \]

### Step 3: Determine the critical points
The critical points occur when:
\[ \sin \theta = 0 \quad \text{or} \quad 2 \sin \theta - 1 = 0 \implies \sin \theta = \frac{1}{2} \]
Thus, the critical points are:
- \( \sin \theta = 0 \) at \( \theta = 0, \pi, 2\pi, \ldots \)
- \( \sin \theta = \frac{1}{2} \) at \( \theta = \frac{\pi}{6}, \frac{5\pi}{6} \)

### Step 4: Analyze the intervals
We analyze the sign of the expression \( -\sin \theta (2 \sin \theta - 1) \) in the intervals determined by the critical points:
- For \( \theta \) in \( (0, \frac{\pi}{6}) \): both factors are negative, so the product is positive.
- For \( \theta \) in \( (\frac{\pi}{6}, \frac{5\pi}{6}) \): \( -\sin \theta \) is negative and \( 2 \sin \theta - 1 \) is positive, so the product is negative.
- For \( \theta \) in \( (\frac{5\pi}{6}, \pi) \): both factors are negative, so the product is positive.
- For \( \theta \) in \( (\pi, \frac{3\pi}{2}) \): \( -\sin \theta \) is negative and \( 2 \sin \theta - 1 \) is negative, so the product is negative.
- For \( \theta \) in \( (\frac{3\pi}{2}, 2\pi) \): both factors are negative, so the product is positive.

From this analysis, we find that the solution to the inequality \( -\sin \theta (2 \sin \theta - 1) \leq 0 \) is:
\[ \theta \in \left[ \frac{\pi}{6}, \frac{5\pi}{6} \right] \cup \left[ \pi, \frac{3\pi}{2} \right] \]

### Step 5: Find the intersection \( A \cap B \)
Now we need to find the intersection of \( A \) and \( B \):
- \( B = \left[ \frac{\pi}{2}, \frac{3\pi}{2} \right] \)
- The intervals from \( A \) are \( \left[ \frac{\pi}{6}, \frac{5\pi}{6} \right] \) and \( \left[ \pi, \frac{3\pi}{2} \right] \).

The intersection is:
\[ A \cap B = \left[ \pi, \frac{3\pi}{2} \right] \]

### Final Answer
Thus, the value of \( A \cap B \) is:
\[ \boxed{\left[ \pi, \frac{3\pi}{2} \right]} \]