The value of x satisfying the equation `cos^(-1)3x+sin^(-1)2x=pi` is

Options

A. `x=(1)/(sqrt(3))`
B. `x=(-1)/(sqrt(3))`
C. `x=(-1)/(sqrt(3))`
D. none of these

Explanation

To solve the equation \( \cos^{-1}(3x) + \sin^{-1}(2x) = \pi \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos^{-1}(3x) + \sin^{-1}(2x) = \pi \]
We can express \( \sin^{-1}(2x) \) in terms of \( \cos^{-1}(3x) \):
\[ \sin^{-1}(2x) = \pi - \cos^{-1}(3x) \]

### Step 2: Use the identity for sine
Using the identity \( \sin(\pi - \theta) = \sin(\theta) \), we can write:
\[ 2x = \sin(\pi - \cos^{-1}(3x)) = \sin(\cos^{-1}(3x)) \]

### Step 3: Express sine in terms of cosine
From the identity \( \sin(\cos^{-1}(y)) = \sqrt{1 - y^2} \), we have:
\[ \sin(\cos^{-1}(3x)) = \sqrt{1 - (3x)^2} = \sqrt{1 - 9x^2} \]
Thus, we can rewrite the equation as:
\[ 2x = \sqrt{1 - 9x^2} \]

### Step 4: Square both sides
To eliminate the square root, we square both sides:
\[ (2x)^2 = (1 - 9x^2) \]
This simplifies to:
\[ 4x^2 = 1 - 9x^2 \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ 4x^2 + 9x^2 = 1 \]
\[ 13x^2 = 1 \]

### Step 6: Solve for \( x^2 \)
Dividing both sides by 13, we find:
\[ x^2 = \frac{1}{13} \]

### Step 7: Find \( x \)
Taking the square root of both sides, we get:
\[ x = \pm \frac{1}{\sqrt{13}} \]

### Step 8: Determine valid solutions
Since \( \cos^{-1}(3x) \) and \( \sin^{-1}(2x) \) must be defined, we check the ranges:
- For \( \cos^{-1}(3x) \) to be defined, \( |3x| \leq 1 \) implies \( |x| \leq \frac{1}{3} \).
- For \( \sin^{-1}(2x) \) to be defined, \( |2x| \leq 1 \) implies \( |x| \leq \frac{1}{2} \).

Since \( \frac{1}{\sqrt{13}} \approx 0.277 \) is less than both \( \frac{1}{3} \) and \( \frac{1}{2} \), both solutions \( x = \frac{1}{\sqrt{13}} \) and \( x = -\frac{1}{\sqrt{13}} \) are valid.

### Final Answer
Thus, the values of \( x \) satisfying the equation are:
\[ x = \pm \frac{1}{\sqrt{13}} \]

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