Let `S` be the set of all solutions of the equation `cos^(-1)(2x)-2cos^(-1)(sqrt(1-x^(2)))=pi,x in[-(1)/(2),(1)/(2)]`.Then `sum_(x=S)2sin^(-1)(x^(2)-1)` is equal to

Options

A. `0`
B. `(-2 pi)/(3)`
C. `pi-2sin^(-1)((sqrt(3))/(4))`
D. `pi-sin^(-1)((sqrt(3))/(4))`

Answer

To solve the equation \( \cos^{-1}(2x) - 2\cos^{-1}(\sqrt{1-x^2}) = \pi \) for \( x \) in the interval \( \left[-\frac{1}{2}, \frac{1}{2}\right] \), we will follow these steps:

### Step 1: Rewrite the equation
We start by rewriting the equation:
\[ \cos^{-1}(2x) - 2\cos^{-1}(\sqrt{1-x^2}) = \pi \]
This can be rearranged to:
\[ \cos^{-1}(2x) = \pi + 2\cos^{-1}(\sqrt{1-x^2}) \]

### Step 2: Apply the cosine function
Taking the cosine of both sides, we have:
\[ 2x = \cos(\pi + 2\cos^{-1}(\sqrt{1-x^2})) \]
Using the identity \( \cos(\pi + \theta) = -\cos(\theta) \):
\[ 2x = -\cos(2\cos^{-1}(\sqrt{1-x^2})) \]

### Step 3: Simplify using cosine double angle formula
Using the double angle formula \( \cos(2\theta) = 2\cos^2(\theta) - 1 \):
\[ \cos(2\cos^{-1}(\sqrt{1-x^2})) = 2(\sqrt{1-x^2})^2 - 1 = 2(1-x^2) - 1 = 1 - 2x^2 \]
Substituting this back, we have:
\[ 2x = -(1 - 2x^2) \]
This simplifies to:
\[ 2x = -1 + 2x^2 \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ 2x^2 - 2x - 1 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = -2, c = -1 \):
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{2 \pm \sqrt{4 + 8}}{4} = \frac{2 \pm \sqrt{12}}{4} = \frac{2 \pm 2\sqrt{3}}{4} = \frac{1 \pm \sqrt{3}}{2} \]

### Step 6: Determine valid solutions within the interval
The solutions are:
\[ x = \frac{1 + \sqrt{3}}{2} \quad \text{and} \quad x = \frac{1 - \sqrt{3}}{2} \]
We check which of these values lie within the interval \( \left[-\frac{1}{2}, \frac{1}{2}\right] \):
- \( \frac{1 + \sqrt{3}}{2} > \frac{1}{2} \) (not valid)
- \( \frac{1 - \sqrt{3}}{2} \) is valid since \( \sqrt{3} \approx 1.732 \) gives \( \frac{1 - 1.732}{2} < -\frac{1}{2} \).

Thus, the only valid solution is:
\[ x = \frac{1 - \sqrt{3}}{2} \]

### Step 7: Calculate the required sum
Now we need to compute:
\[ \sum_{x \in S} 2\sin^{-1}(x^2 - 1) \]
Calculating \( x^2 \):
\[ x^2 = \left(\frac{1 - \sqrt{3}}{2}\right)^2 = \frac{(1 - 2\sqrt{3} + 3)}{4} = \frac{4 - 2\sqrt{3}}{4} = 1 - \frac{\sqrt{3}}{2} \]
Thus:
\[ x^2 - 1 = -\frac{\sqrt{3}}{2} \]
Now, calculate \( 2\sin^{-1}(-\frac{\sqrt{3}}{2}) \):
\[ \sin^{-1}(-\frac{\sqrt{3}}{2}) = -\frac{\pi}{3} \]
Therefore:
\[ 2\sin^{-1}(-\frac{\sqrt{3}}{2}) = 2 \cdot -\frac{\pi}{3} = -\frac{2\pi}{3} \]

### Final Answer
The final answer is:
\[ \boxed{-\frac{2\pi}{3}} \]