Statement I `sin^(-1) 2x + sin^(-1) 3x = pi/3 `
`rArr x = sqrt(3/76)` only.
and
Statement II Sum of twp negative angles cannot be positive.

Options

A. Statement I is True, Statement II is True, Statement II is a correct explanation for statement I
B. Statement I is True, Statement II is True, Statement II is NOT a correct explanation for Statement I
C. Statement I is True, Statement II is False
D. Statement I is False, Statement II is True.

Answer

To solve the problem, we need to analyze the given statements and verify their validity step by step.

### Step 1: Analyze Statement I
We need to solve the equation:
\[ \sin^{-1}(2x) + \sin^{-1}(3x) = \frac{\pi}{3} \]

### Step 2: Apply the Formula for Sine Inverse
We will use the formula for the sum of inverse sine functions:
\[ \sin^{-1}(a) + \sin^{-1}(b) = \sin^{-1}\left(a\sqrt{1-b^2} + b\sqrt{1-a^2}\right) \]
Here, let \( a = 2x \) and \( b = 3x \).

### Step 3: Substitute into the Formula
Substituting \( a \) and \( b \):
\[ \sin^{-1}(2x) + \sin^{-1}(3x) = \sin^{-1}\left(2x \sqrt{1 - (3x)^2} + 3x \sqrt{1 - (2x)^2}\right) \]
This gives us:
\[ \sin^{-1}\left(2x \sqrt{1 - 9x^2} + 3x \sqrt{1 - 4x^2}\right) = \frac{\pi}{3} \]

### Step 4: Take Sine of Both Sides
Taking the sine of both sides:
\[ 2x \sqrt{1 - 9x^2} + 3x \sqrt{1 - 4x^2} = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]

### Step 5: Rearranging the Equation
Rearranging gives us:
\[ 2x \sqrt{1 - 9x^2} + 3x \sqrt{1 - 4x^2} - \frac{\sqrt{3}}{2} = 0 \]

### Step 6: Substitute \( x = \sqrt{\frac{3}{76}} \)
Now, we will substitute \( x = \sqrt{\frac{3}{76}} \) into the equation to verify if it holds true.

Calculating \( 9x^2 \) and \( 4x^2 \):
\[ 9x^2 = 9 \cdot \frac{3}{76} = \frac{27}{76}, \quad 4x^2 = 4 \cdot \frac{3}{76} = \frac{12}{76} \]
Thus,
\[ 1 - 9x^2 = 1 - \frac{27}{76} = \frac{49}{76}, \quad 1 - 4x^2 = 1 - \frac{12}{76} = \frac{64}{76} \]

### Step 7: Substitute Back into the Equation
Now substituting back:
\[ 2\sqrt{\frac{3}{76}} \sqrt{\frac{49}{76}} + 3\sqrt{\frac{3}{76}} \sqrt{\frac{64}{76}} = \frac{\sqrt{3}}{2} \]
Calculating each term:
\[ 2\sqrt{\frac{3 \cdot 49}{76^2}} + 3\sqrt{\frac{3 \cdot 64}{76^2}} = \frac{\sqrt{3}}{2} \]
This simplifies to:
\[ \frac{2\sqrt{3 \cdot 49} + 3\sqrt{3 \cdot 64}}{76} = \frac{\sqrt{3}}{2} \]

### Step 8: Verify the Equality
Now we check if both sides are equal. After simplification, we find that both sides indeed equal, confirming that Statement I is true.

### Step 9: Analyze Statement II
Statement II states that the sum of two negative angles cannot be positive. This is a true statement since the sum of two negative values will always yield a negative value.

### Conclusion
Both statements are true:
- Statement I is true.
- Statement II is true.