If `a_(i)in(-pi,2pi)" for "i=2,3,...,n,` then the number of solutions of the inequailty
`2^(1//sin^(2)a_(2)).3^(1//sin^(2)a_(3)).4^(1//sin^(2)a_(4))...n^(1//sin^(2)a_(n))len!,` is

Options

A. infinite
B. `2^(n-1)`
C. `3^(n-1)`
D. `3^(n)`

Explanation

To solve the inequality

\[ 2^{\frac{1}{\sin^2 a_2}} \cdot 3^{\frac{1}{\sin^2 a_3}} \cdots n^{\frac{1}{\sin^2 a_n}} \leq n! \]

where \( a_i \in (-\pi, 2\pi) \) for \( i = 2, 3, \ldots, n \), we will proceed step by step.

### Step 1: Understand the Range of \( \sin^2 a_i \)

Since \( a_i \) is in the interval \( (-\pi, 2\pi) \), we know that \( \sin^2 a_i \) can take values between 0 and 1. However, for the terms \( \frac{1}{\sin^2 a_i} \) to be defined and valid, \( \sin^2 a_i \) must be greater than 0. Therefore, we need to consider the points where \( \sin a_i = 0 \), which are \( a_i = k\pi \) for \( k \in \mathbb{Z} \).

### Step 2: Identify the Values of \( a_i \) Where \( \sin^2 a_i = 1 \)

The maximum value of \( \sin^2 a_i \) is 1, which occurs at:

\[ a_i = \frac{\pi}{2}, \quad a_i = \frac{3\pi}{2} \]

These values correspond to \( \sin^2 a_i = 1 \).

### Step 3: Set Up the Inequality

For the inequality to hold, we need:

\[ 2^{\frac{1}{\sin^2 a_2}} \cdot 3^{\frac{1}{\sin^2 a_3}} \cdots n^{\frac{1}{\sin^2 a_n}} \leq n! \]

This implies that each term \( i^{\frac{1}{\sin^2 a_i}} \) should be minimized. The minimum occurs when \( \sin^2 a_i = 1 \), giving:

\[ i^{\frac{1}{\sin^2 a_i}} = i^1 = i \]

Thus, the left-hand side becomes:

\[ 2 \cdot 3 \cdots n = n! \]

### Step 4: Determine the Conditions for Equality

The equality holds when \( \sin^2 a_i = 1 \) for all \( i \). This means:

\[ a_i = \frac{\pi}{2} \quad \text{or} \quad a_i = \frac{3\pi}{2} \]

### Step 5: Count the Number of Solutions

For each \( a_i \) (where \( i = 2, 3, \ldots, n \)), there are 2 possible values (\(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\)). Therefore, the total number of combinations for \( n-1 \) values (since \( i \) starts from 2) is:

\[ 2^{n-1} \]

### Final Answer

Thus, the total number of solutions to the inequality is:

\[ \text{Total Solutions} = 2^{n-1} \]