Solve `2 cos^(2) x+ sin x le 2`, where `pi//2 le x le 3pi//2`.

Explanation

To solve the inequality \(2 \cos^2 x + \sin x \leq 2\) for \(x\) in the interval \(\left[\frac{3\pi}{2}, \frac{5\pi}{2}\right]\), we can follow these steps:

### Step 1: Rewrite the inequality
We start with the given inequality:
\[ 2 \cos^2 x + \sin x \leq 2 \]
Using the Pythagorean identity \(\cos^2 x = 1 - \sin^2 x\), we can rewrite \(\cos^2 x\):
\[ 2(1 - \sin^2 x) + \sin x \leq 2 \]
This simplifies to:
\[ 2 - 2\sin^2 x + \sin x \leq 2 \]

### Step 2: Simplify the inequality
Subtract \(2\) from both sides:
\[ -2\sin^2 x + \sin x \leq 0 \]
Rearranging gives:
\[ 2\sin^2 x - \sin x \geq 0 \]

### Step 3: Factor the expression
Factoring out \(\sin x\):
\[ \sin x(2\sin x - 1) \geq 0 \]

### Step 4: Find critical points
The critical points occur when:
\[ \sin x = 0 \quad \text{or} \quad 2\sin x - 1 = 0 \]
Solving \(2\sin x - 1 = 0\) gives:
\[ \sin x = \frac{1}{2} \]

### Step 5: Determine the intervals
The critical points in the interval \(\left[\frac{3\pi}{2}, \frac{5\pi}{2}\right]\) are:
- \(\sin x = 0\) at \(x = 2\pi\) (within the interval)
- \(\sin x = \frac{1}{2}\) at \(x = \frac{7\pi}{6}\) and \(x = \frac{11\pi}{6}\) (but only \(\frac{7\pi}{6}\) is in the interval)

### Step 6: Test the intervals
We need to test the intervals determined by the critical points:
1. \(\left[\frac{3\pi}{2}, \frac{7\pi}{6}\right)\)
2. \(\left[\frac{7\pi}{6}, 2\pi\right)\)
3. \(\left[2\pi, \frac{5\pi}{2}\right]\)

- For \(x = \frac{3\pi}{2}\):
\(\sin\left(\frac{3\pi}{2}\right) = -1\), so \(\sin x(2\sin x - 1) < 0\).

- For \(x = \frac{7\pi}{6}\):
\(\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}\), so \(\sin x(2\sin x - 1) < 0\).

- For \(x = 2\pi\):
\(\sin(2\pi) = 0\), so \(\sin x(2\sin x - 1) = 0\).

- For \(x = \frac{5\pi}{2}\):
\(\sin\left(\frac{5\pi}{2}\right) = 1\), so \(\sin x(2\sin x - 1) > 0\).

### Step 7: Conclusion
The solution to the inequality \(2 \cos^2 x + \sin x \leq 2\) in the interval \(\left[\frac{3\pi}{2}, \frac{5\pi}{2}\right]\) is:
\[ x \in \left[2\pi, \frac{5\pi}{2}\right] \]