The solution of `sin^(-1)x-sin^(-1)2x=pm(pi)/(3)` is

Options

A. `pm(1)/(3)`
B. `pm(1)/(4)`
C. `pm(sqrt(3))/(2)`
D. `pm(1)/(2)`

Explanation

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

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin^{-1} x - \sin^{-1} 2x = \pm \frac{\pi}{3} \]
We can express \( \frac{\pi}{3} \) in terms of sine:
\[ \frac{\pi}{3} = \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) \]
Thus, we can rewrite the equation as:
\[ \sin^{-1} x - \sin^{-1} 2x = \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) \quad \text{or} \quad \sin^{-1} x - \sin^{-1} 2x = -\sin^{-1} \left( \frac{\sqrt{3}}{2} \right) \]

### Step 2: Apply the sine difference formula
Using the identity for the difference of inverse sine functions:
\[ \sin^{-1} a - \sin^{-1} b = \sin^{-1 \left( a \sqrt{1 - b^2} - b \sqrt{1 - a^2} \right)} \]
we can set \( a = x \) and \( b = 2x \):
\[ \sin^{-1} x - \sin^{-1} 2x = \sin^{-1} \left( x \sqrt{1 - (2x)^2} - 2x \sqrt{1 - x^2} \right) \]

### Step 3: Set the expressions equal
Now we have:
\[ \sin^{-1} \left( x \sqrt{1 - 4x^2} - 2x \sqrt{1 - x^2} \right) = \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) \]
This implies:
\[ x \sqrt{1 - 4x^2} - 2x \sqrt{1 - x^2} = \frac{\sqrt{3}}{2} \]
or
\[ x \sqrt{1 - 4x^2} - 2x \sqrt{1 - x^2} = -\frac{\sqrt{3}}{2} \]

### Step 4: Solve the first equation
Let’s solve the first equation:
\[ x \sqrt{1 - 4x^2} - 2x \sqrt{1 - x^2} = \frac{\sqrt{3}}{2} \]
Rearranging gives:
\[ x \sqrt{1 - 4x^2} = \frac{\sqrt{3}}{2} + 2x \sqrt{1 - x^2} \]
Squaring both sides:
\[ x^2 (1 - 4x^2) = \left(\frac{\sqrt{3}}{2} + 2x \sqrt{1 - x^2}\right)^2 \]

### Step 5: Expand and simplify
Expanding the right side:
\[ x^2 - 4x^4 = \frac{3}{4} + 2\sqrt{3}x\sqrt{1 - x^2} + 4x^2(1 - x^2) \]
This leads to a quadratic equation in \( x \).

### Step 6: Solve the quadratic equation
After simplification, we will arrive at a quadratic equation:
\[ 4x^4 + 3x^2 - 1 = 0 \]
Let \( y = x^2 \):
\[ 4y^2 + 3y - 1 = 0 \]
Using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} \]
Calculating gives:
\[ y = \frac{-3 \pm \sqrt{9 + 16}}{8} = \frac{-3 \pm 5}{8} \]
Thus, \( y = \frac{1}{4} \) or \( y = -1 \) (not valid since \( y = x^2 \geq 0 \)).

### Step 7: Find values of \( x \)
Since \( y = x^2 = \frac{1}{4} \):
\[ x = \pm \frac{1}{2} \]

### Final Answer
The solutions to the original equation are:
\[ x = \frac{1}{2} \quad \text{and} \quad x = -\frac{1}{2} \]