If `theta in [0,5pi]a n dr in R` such that `2sintheta=r^4-2r^2+3` then the maximum number of values of the pair `(r ,theta)` is____

Explanation

To solve the equation \( 2 \sin \theta = r^4 - 2r^2 + 3 \) and find the maximum number of values of the pair \( (r, \theta) \), we can follow these steps:

### Step 1: Rearranging the Equation
Start with the given equation:
\[ 2 \sin \theta = r^4 - 2r^2 + 3 \]
We can rearrange it to express \( \sin \theta \):
\[ \sin \theta = \frac{r^4 - 2r^2 + 3}{2} \]

### Step 2: Analyzing the Right-Hand Side
Next, we need to analyze the expression \( r^4 - 2r^2 + 3 \). Let's denote:
\[ f(r) = r^4 - 2r^2 + 3 \]
To understand the behavior of this function, we can find its minimum value.

### Step 3: Finding the Minimum Value of \( f(r) \)
To find the minimum value of \( f(r) \), we can take the derivative and set it to zero:
\[ f'(r) = 4r^3 - 4r = 4r(r^2 - 1) \]
Setting \( f'(r) = 0 \) gives us:
\[ 4r(r^2 - 1) = 0 \]
Thus, \( r = 0, r = 1, r = -1 \).

### Step 4: Evaluating \( f(r) \) at Critical Points
Now, we evaluate \( f(r) \) at these critical points:
1. \( f(0) = 0^4 - 2(0^2) + 3 = 3 \)
2. \( f(1) = 1^4 - 2(1^2) + 3 = 1 - 2 + 3 = 2 \)
3. \( f(-1) = (-1)^4 - 2(-1)^2 + 3 = 1 - 2 + 3 = 2 \)

### Step 5: Finding the Minimum Value
The minimum value of \( f(r) \) is \( 2 \) (occurring at \( r = 1 \) and \( r = -1 \)). Therefore, we have:
\[ f(r) \geq 2 \text{ for all } r \in \mathbb{R} \]

### Step 6: Setting Up the Sine Condition
Since \( \sin \theta \) must be in the range \([-1, 1]\), we need:
\[ \frac{r^4 - 2r^2 + 3}{2} \leq 1 \]
This simplifies to:
\[ r^4 - 2r^2 + 3 \leq 2 \implies r^4 - 2r^2 + 1 \leq 0 \]
Factoring gives:
\[ (r^2 - 1)^2 \leq 0 \]
The only solution to this is when \( r^2 - 1 = 0 \), which means:
\[ r^2 = 1 \implies r = 1 \text{ or } r = -1 \]

### Step 7: Finding Corresponding Values of \( \theta \)
Now, we need to find the values of \( \theta \) when \( r = 1 \) or \( r = -1 \):
1. For \( r = 1 \):
\[ 2 \sin \theta = 2 \implies \sin \theta = 1 \implies \theta = \frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2} \]
2. For \( r = -1 \):
\[ 2 \sin \theta = 2 \implies \sin \theta = 1 \implies \theta = \frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2} \]

### Step 8: Counting Unique Pairs
In total, the values of \( \theta \) corresponding to both \( r = 1 \) and \( r = -1 \) are:
- \( \frac{\pi}{2} \)
- \( \frac{5\pi}{2} \)
- \( \frac{9\pi}{2} \)

Thus, we have 3 unique values of \( \theta \) for each \( r \) value, leading to:
\[ \text{Total pairs } (r, \theta) = 3 \text{ (for } r = 1) + 3 \text{ (for } r = -1) = 6 \]

### Final Answer
The maximum number of values of the pair \( (r, \theta) \) is:
\[ \boxed{6} \]