Find the set of values of `alpha` in the interval [ `pi/2,3pi/2`], for which the point (`sin alpha, cos alpha`)does not exist outside the parabola `2 y^2 + x - 2 = 0`

Options

A. `[pi//2, 5pi//6]`
B. `[pi, 3pi//2]`
C. `[pi//2, 5pi//6]uu[pi,3pi//2]`
D. `[5pi//6, 3pi//2]`

Explanation

To solve the problem of finding the set of values of \( \alpha \) in the interval \([ \frac{\pi}{2}, \frac{3\pi}{2} ]\) for which the point \((\sin \alpha, \cos \alpha)\) does not exist outside the parabola \(2y^2 + x - 2 = 0\), we will follow these steps:

### Step 1: Rewrite the parabola equation
The given equation of the parabola is:
\[ 2y^2 + x - 2 = 0 \]
We can rewrite this as:
\[ x = 2 - 2y^2 \]
This shows that the parabola opens to the left.

### Step 2: Substitute the point into the parabola equation
We need to check when the point \((\sin \alpha, \cos \alpha)\) lies inside or on the parabola. Substituting \(x = \sin \alpha\) and \(y = \cos \alpha\) into the parabola equation gives:
\[ 2(\cos \alpha)^2 + \sin \alpha - 2 \leq 0 \]

### Step 3: Simplify the inequality
Using the identity \(\cos^2 \alpha = 1 - \sin^2 \alpha\), we can rewrite the inequality:
\[ 2(1 - \sin^2 \alpha) + \sin \alpha - 2 \leq 0 \]
This simplifies to:
\[ 2 - 2\sin^2 \alpha + \sin \alpha - 2 \leq 0 \]
Which further simplifies to:
\[ -2\sin^2 \alpha + \sin \alpha \leq 0 \]

### Step 4: Factor the inequality
Factoring out \(-\sin \alpha\) gives:
\[ -\sin \alpha(2\sin \alpha - 1) \leq 0 \]
This means that either \(\sin \alpha = 0\) or \(2\sin \alpha - 1 = 0\).

### Step 5: Solve the inequalities
1. From \(\sin \alpha = 0\):
- In the interval \([ \frac{\pi}{2}, \frac{3\pi}{2} ]\), \(\sin \alpha = 0\) at \(\alpha = \pi\).

2. From \(2\sin \alpha - 1 = 0\):
- This gives \(\sin \alpha = \frac{1}{2}\).
- In the interval \([ \frac{\pi}{2}, \frac{3\pi}{2} ]\), \(\sin \alpha = \frac{1}{2}\) at \(\alpha = \frac{5\pi}{6}\).

### Step 6: Determine the intervals
Now we need to check the intervals:
- The critical points are \(\alpha = \pi\) and \(\alpha = \frac{5\pi}{6}\).
- We check the signs of the expression \(-\sin \alpha(2\sin \alpha - 1)\) in the intervals:
- For \(\alpha \in [\frac{\pi}{2}, \frac{5\pi}{6}]\), \(\sin \alpha\) is positive and \(2\sin \alpha - 1\) is negative, so the product is negative.
- For \(\alpha = \frac{5\pi}{6}\), the product is zero.
- For \(\alpha \in [\frac{5\pi}{6}, \pi]\), the product is zero at \(\frac{5\pi}{6}\) and negative until \(\pi\).
- For \(\alpha \in [\pi, \frac{3\pi}{2}]\), \(\sin \alpha\) is zero at \(\pi\) and negative afterwards.

### Step 7: Final intervals
Thus, the values of \(\alpha\) for which the point \((\sin \alpha, \cos \alpha)\) does not exist outside the parabola are:
\[ \alpha \in \left[\frac{\pi}{2}, \frac{5\pi}{6}\right] \cup \left[\pi, \frac{3\pi}{2}\right] \]

### Final Answer
The set of values of \( \alpha \) is:
\[ \left[\frac{\pi}{2}, \frac{5\pi}{6}\right] \cup \left[\pi, \frac{3\pi}{2}\right] \]