If `lim_(xto0) [1+xln(1+b^(2))]^(1//x)=2bsin^(2)theta,bgt0," and "thetain(-pi,pi],` then the value of `theta` is

Explanation

To solve the limit problem, we start with the expression given in the limit:

\[ \lim_{x \to 0} \left[1 + x \ln(1 + b^2)\right]^{\frac{1}{x}} = 2b \sin^2 \theta \]

### Step 1: Recognize the form of the limit
As \( x \to 0 \), the expression \( 1 + x \ln(1 + b^2) \) approaches \( 1 \), and we have a limit of the form \( 1^{\infty} \). To evaluate this limit, we can use the exponential limit property:

\[ \lim_{x \to 0} \left[1 + f(x)\right]^{g(x)} = e^{\lim_{x \to 0} f(x) g(x)} \]

where \( f(x) = x \ln(1 + b^2) \) and \( g(x) = \frac{1}{x} \).

### Step 2: Calculate \( f(x) g(x) \)
We have:

\[ f(x) g(x) = x \ln(1 + b^2) \cdot \frac{1}{x} = \ln(1 + b^2) \]

Thus, we can rewrite the limit as:

\[ \lim_{x \to 0} \left[1 + x \ln(1 + b^2)\right]^{\frac{1}{x}} = e^{\ln(1 + b^2)} = 1 + b^2 \]

### Step 3: Set the limit equal to the right side
We equate the limit to the right-hand side of the original equation:

\[ 1 + b^2 = 2b \sin^2 \theta \]

### Step 4: Rearranging the equation
Rearranging gives us:

\[ b^2 - 2b \sin^2 \theta + 1 = 0 \]

### Step 5: Solve the quadratic equation
This is a quadratic equation in terms of \( b \). The discriminant must be non-negative for real solutions:

\[ D = (2 \sin^2 \theta)^2 - 4 \cdot 1 \cdot 1 = 4 \sin^4 \theta - 4 \]

Setting the discriminant \( D \geq 0 \):

\[ 4(\sin^4 \theta - 1) \geq 0 \]

This simplifies to:

\[ \sin^4 \theta - 1 \geq 0 \]

### Step 6: Factor the inequality
Factoring gives us:

\[ (\sin^2 \theta - 1)(\sin^2 \theta + 1) \geq 0 \]

Since \( \sin^2 \theta + 1 > 0 \) for all \( \theta \), we focus on:

\[ \sin^2 \theta - 1 \geq 0 \]

This implies:

\[ \sin^2 \theta \geq 1 \]

### Step 7: Find values of \( \theta \)
The only values of \( \theta \) that satisfy \( \sin^2 \theta = 1 \) are:

\[ \theta = \frac{\pi}{2} \quad \text{or} \quad \theta = -\frac{\pi}{2} \]

### Conclusion
Thus, the values of \( \theta \) that satisfy the original limit condition are:

\[ \theta = \frac{\pi}{2} \quad \text{and} \quad \theta = -\frac{\pi}{2} \]