The number of solutions of `sin^(-1) x+ sin^(-1) 2x=pi/3` is

Options

A. 0
B. 1
C. 2
D. infinite

Answer

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

### Step 1: Rewrite the Equation
We start with the equation:
\[ \sin^{-1} x + \sin^{-1} 2x = \frac{\pi}{3} \]
We can rewrite this as:
\[ \sin^{-1} 2x = \frac{\pi}{3} - \sin^{-1} x \]

### Step 2: Apply the Sine Function
Taking the sine of both sides, we have:
\[ 2x = \sin\left(\frac{\pi}{3} - \sin^{-1} x\right) \]

### Step 3: Use the Sine Subtraction Formula
Using the sine subtraction formula, we can express the right-hand side:
\[ \sin\left(\frac{\pi}{3} - \sin^{-1} x\right) = \sin\frac{\pi}{3} \cos(\sin^{-1} x) - \cos\frac{\pi}{3} \sin(\sin^{-1} x) \]
We know that \( \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2} \) and \( \cos\frac{\pi}{3} = \frac{1}{2} \). Also, \( \sin(\sin^{-1} x) = x \) and \( \cos(\sin^{-1} x) = \sqrt{1 - x^2} \). Therefore, we have:
\[ 2x = \frac{\sqrt{3}}{2} \sqrt{1 - x^2} - \frac{1}{2} x \]

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

### Step 5: Squaring Both Sides
Squaring both sides to eliminate the square root:
\[ \left(\frac{5}{2} x\right)^2 = \left(\frac{\sqrt{3}}{2} \sqrt{1 - x^2}\right)^2 \]
This leads to:
\[ \frac{25}{4} x^2 = \frac{3}{4} (1 - x^2) \]

### Step 6: Simplifying the Equation
Multiplying through by 4 to eliminate the denominators:
\[ 25x^2 = 3(1 - x^2) \]
Expanding gives:
\[ 25x^2 = 3 - 3x^2 \]
Combining like terms results in:
\[ 28x^2 = 3 \]

### Step 7: Solving for x
Dividing both sides by 28:
\[ x^2 = \frac{3}{28} \]
Taking the square root gives:
\[ x = \pm \sqrt{\frac{3}{28}} = \pm \frac{\sqrt{3}}{2\sqrt{7}} = \pm \frac{\sqrt{3}}{2\sqrt{4 \cdot 7}} = \pm \frac{\sqrt{3}}{2 \cdot 2\sqrt{7}} = \pm \frac{\sqrt{3}}{4\sqrt{7}} \]

### Step 8: Validating the Solutions
Since \( \sin^{-1} x \) and \( \sin^{-1} 2x \) are defined only for \( x \) in the interval \([-1, 1]\), we need to check which of these values are valid. The positive value will be valid, while the negative value will not yield a valid solution since both \( \sin^{-1} x \) and \( \sin^{-1} 2x \) would yield negative results, which cannot sum to a positive angle like \( \frac{\pi}{3} \).

### Conclusion
Thus, the number of solutions to the equation \( \sin^{-1} x + \sin^{-1} 2x = \frac{\pi}{3} \) is:
\[ \text{Number of solutions} = 1 \]