The value of x satisfying `sin^(-1)(sqrt((3x-1)/(25)))+sin^(-1)(sqrt((3x+1)/(25)))=(pi)/(2)` lies in the interval

Options

A. (1,2)
B. (2,3)
C. (3,4)
D. (4,5)

Explanation

To solve the equation

\[ \sin^{-1}\left(\sqrt{\frac{3x-1}{25}}\right) + \sin^{-1}\left(\sqrt{\frac{3x+1}{25}}\right) = \frac{\pi}{2}, \]

we can use the property of inverse trigonometric functions that states:

\[ \sin^{-1}(a) + \cos^{-1}(a) = \frac{\pi}{2}. \]

### Step 1: Rewrite the equation using the property

Using the property mentioned, we can rewrite the equation as:

\[ \sin^{-1}\left(\sqrt{\frac{3x-1}{25}}\right) = \cos^{-1}\left(\sqrt{\frac{3x+1}{25}}\right). \]

### Step 2: Convert cosine inverse to sine inverse

We can convert the cosine inverse to sine inverse:

\[ \sin^{-1}\left(\sqrt{\frac{3x-1}{25}}\right) = \sin^{-1}\left(\sqrt{1 - \left(\sqrt{\frac{3x+1}{25}}\right)^2}\right). \]

### Step 3: Simplify the right-hand side

Now, we simplify the right-hand side:

\[ \sqrt{1 - \left(\sqrt{\frac{3x+1}{25}}\right)^2} = \sqrt{1 - \frac{3x+1}{25}} = \sqrt{\frac{25 - (3x + 1)}{25}} = \sqrt{\frac{24 - 3x}{25}}. \]

### Step 4: Set the two sides equal

Now we have:

\[ \sqrt{\frac{3x-1}{25}} = \sqrt{\frac{24 - 3x}{25}}. \]

### Step 5: Square both sides

Squaring both sides gives:

\[ \frac{3x-1}{25} = \frac{24 - 3x}{25}. \]

### Step 6: Eliminate the denominator

Multiplying both sides by 25:

\[ 3x - 1 = 24 - 3x. \]

### Step 7: Solve for x

Now, we can solve for \(x\):

\[ 3x + 3x = 24 + 1, \]
\[ 6x = 25, \]
\[ x = \frac{25}{6}. \]

### Step 8: Determine the interval

Now we need to check if this value of \(x\) lies in the specified interval.

Calculating \( \frac{25}{6} \):

\[ \frac{25}{6} \approx 4.1667. \]

This value lies in the interval \( (4, 5) \).

### Final Answer

Thus, the value of \(x\) satisfying the equation lies in the interval \( (4, 5) \).

---