Number of solutions of the equation `csctheta=k` in `[0,pi]` where `k=lim_(n->oo)prod_(r=2)^n((r^3-1)/(r^3+1))` is ____

Answer

To solve the problem step by step, we need to find the value of \( k \) and then determine the number of solutions to the equation \( \csc \theta = k \) in the interval \([0, \pi]\).

### Step 1: Calculate \( k \)

We start with the expression for \( k \):

\[ k = \lim_{n \to \infty} \prod_{r=2}^n \frac{r^3 - 1}{r^3 + 1} \]

### Step 2: Simplify the product

We can rewrite the fraction inside the product:

\[ \frac{r^3 - 1}{r^3 + 1} = \frac{(r - 1)(r^2 + r + 1)}{(r + 1)(r^2 - r + 1)} \]

Thus, we can express \( k \) as:

\[ k = \lim_{n \to \infty} \left( \prod_{r=2}^n \frac{r - 1}{r + 1} \cdot \prod_{r=2}^n \frac{r^2 + r + 1}{r^2 - r + 1} \right) \]

Let’s denote these products as \( P_1 \) and \( P_2 \):

\[ P_1 = \prod_{r=2}^n \frac{r - 1}{r + 1}, \quad P_2 = \prod_{r=2}^n \frac{r^2 + r + 1}{r^2 - r + 1} \]

### Step 3: Evaluate \( P_1 \)

For \( P_1 \):

\[ P_1 = \frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \cdots \frac{n-1}{n+1} \]

This can be simplified as:

\[ P_1 = \frac{2 \cdot 1}{n(n+1)} \to 0 \text{ as } n \to \infty \]

### Step 4: Evaluate \( P_2 \)

For \( P_2 \):

\[ P_2 = \prod_{r=2}^n \frac{r^2 + r + 1}{r^2 - r + 1} \]

This product can be evaluated similarly, but as \( n \to \infty \), the contributions of the terms will also converge.

### Step 5: Combine results

As \( n \to \infty \), we find that:

\[ k = \lim_{n \to \infty} \left( P_1 \cdot P_2 \right) = 0 \]

### Step 6: Solve \( \csc \theta = k \)

Now we have:

\[ \csc \theta = 0 \]

### Step 7: Determine the number of solutions

The cosecant function \( \csc \theta \) is defined as \( \frac{1}{\sin \theta} \). The equation \( \csc \theta = 0 \) implies that \( \sin \theta \) must be undefined, which does not occur for any \( \theta \) in the interval \([0, \pi]\).

### Conclusion

Thus, there are no solutions to the equation \( \csc \theta = 0 \) in the interval \([0, \pi]\).

The final answer is:

\[ \text{Number of solutions} = 0 \]