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:
\[ \sin^{-1} x - \sin^{-1} 2x = \pm \frac{\pi}{3} \]

### Step 2: Use the sine subtraction formula
Using the sine subtraction formula, we can express this as:
\[ \sin^{-1} x - \sin^{-1} 2x = \sin^{-1}\left( x \sqrt{1 - (2x)^2} \right) = \pm \frac{\pi}{3} \]
This implies:
\[ x \sqrt{1 - 4x^2} = \sin\left(\pm \frac{\pi}{3}\right) \]

### Step 3: Evaluate \(\sin\left(\pm \frac{\pi}{3}\right)\)
The value of \(\sin\left(\frac{\pi}{3}\right)\) is \(\frac{\sqrt{3}}{2}\). Therefore, we have:
\[ x \sqrt{1 - 4x^2} = \pm \frac{\sqrt{3}}{2} \]

### Step 4: Square both sides
Squaring both sides gives:
\[ x^2 (1 - 4x^2) = \frac{3}{4} \]
This simplifies to:
\[ x^2 - 4x^4 = \frac{3}{4} \]

### Step 5: Rearranging the equation
Rearranging the equation results in:
\[ 4x^4 - x^2 + \frac{3}{4} = 0 \]
Multiplying through by 4 to eliminate the fraction:
\[ 16x^4 - 4x^2 + 3 = 0 \]

### Step 6: Let \(y = x^2\)
Let \(y = x^2\). The equation becomes:
\[ 16y^2 - 4y + 3 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ y = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 16 \cdot 3}}{2 \cdot 16} \]
Calculating the discriminant:
\[ y = \frac{4 \pm \sqrt{16 - 192}}{32} \]
\[ y = \frac{4 \pm \sqrt{-176}}{32} \]
Since the discriminant is negative, there are no real solutions for \(y\).

### Step 8: Conclusion
Since \(y = x^2\) has no real solutions, we conclude that the original equation \( \sin^{-1} x - \sin^{-1} 2x = \pm \frac{\pi}{3} \) has no real solutions.

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