If `f(x)={{:(((2^(x)-1)^(2)tan3x)/(xsin^(2)x)":",0ltxltpi//6),(lambda":",x=0):}` is continuous at x = 0, then the value of `(10sqrt(3lambda))/(ln2)` is equal to

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = 0 \). This means that the limit of \( f(x) \) as \( x \) approaches 0 from the right must equal \( f(0) \), which is given as \( \lambda \).

### Step-by-step Solution:

1. **Identify the function**:
The function is defined as:
\[ f(x) = \frac{(2^x - 1)^2 \tan(3x)}{x \sin^2(x)} \quad \text{for } 0 < x < \frac{\pi}{6} \]
and \( f(0) = \lambda \).

2. **Find the limit as \( x \) approaches 0**:
We need to compute:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{(2^x - 1)^2 \tan(3x)}{x \sin^2(x)} \]

3. **Use Taylor expansions**:
- For small \( x \), we can use the approximations:
\[ 2^x - 1 \approx x \ln(2) \]
\[ \tan(3x) \approx 3x \]
\[ \sin(x) \approx x \]
Therefore, \( \sin^2(x) \approx x^2 \).

4. **Substitute these approximations into the limit**:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{(x \ln(2))^2 \cdot 3x}{x \cdot x^2} \]
Simplifying this gives:
\[ = \lim_{x \to 0^+} \frac{3 (\ln(2))^2 x^3}{x^3} = 3 (\ln(2))^2 \]

5. **Set the limit equal to \( \lambda \)**:
Since \( f(0) = \lambda \), we have:
\[ \lambda = 3 (\ln(2))^2 \]

6. **Calculate \( \frac{10 \sqrt{3} \lambda}{\ln(2)} \)**:
Substituting \( \lambda \) into the expression:
\[ \frac{10 \sqrt{3} \lambda}{\ln(2)} = \frac{10 \sqrt{3} \cdot 3 (\ln(2))^2}{\ln(2)} = 30 \sqrt{3} \ln(2) \]

### Final Answer:
The value of \( \frac{10 \sqrt{3} \lambda}{\ln(2)} \) is \( 30 \sqrt{3} \ln(2) \).