Find the value of `cos^(-1)(x)+cos^(-1)((x)/(2)+(sqrt(3-3x^(2)))/(2))`

Options

A. `(pi)/(6)`
B. `(pi)/(3)`
C. `(pi)/(4)`
D. `(pi)/(12)`

Explanation

To solve the problem, we need to find the value of \( \cos^{-1}(x) + \cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3 - 3x^2}}{2}\right) \).

### Step-by-Step Solution:

1. **Let \( \alpha = \cos^{-1}(x) \)**:
\[ x = \cos(\alpha) \]

2. **Rewrite the second term**:
We need to evaluate \( \cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3 - 3x^2}}{2}\right) \). Substitute \( x = \cos(\alpha) \):
\[ \frac{x}{2} = \frac{\cos(\alpha)}{2} \]
Now, we need to simplify \( \sqrt{3 - 3x^2} \):
\[ \sqrt{3 - 3x^2} = \sqrt{3 - 3\cos^2(\alpha)} = \sqrt{3(1 - \cos^2(\alpha))} = \sqrt{3\sin^2(\alpha)} = \sqrt{3} \sin(\alpha) \]
Therefore, we can rewrite the expression:
\[ \frac{x}{2} + \frac{\sqrt{3 - 3x^2}}{2} = \frac{\cos(\alpha)}{2} + \frac{\sqrt{3} \sin(\alpha)}{2} = \frac{1}{2}(\cos(\alpha) + \sqrt{3} \sin(\alpha)) \]

3. **Recognize the cosine of a sum**:
The expression \( \cos(\alpha) + \sqrt{3} \sin(\alpha) \) can be recognized as:
\[ \cos(\alpha) + \sqrt{3} \sin(\alpha) = 2 \left(\frac{1}{2} \cos(\alpha) + \frac{\sqrt{3}}{2} \sin(\alpha)\right) = 2 \cos\left(\alpha - \frac{\pi}{3}\right) \]
Thus,
\[ \frac{1}{2}(\cos(\alpha) + \sqrt{3} \sin(\alpha)) = \cos\left(\alpha - \frac{\pi}{3}\right) \]

4. **Substituting back**:
Now we have:
\[ \cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3 - 3x^2}}{2}\right) = \cos^{-1}\left(\cos\left(\alpha - \frac{\pi}{3}\right)\right) \]
This simplifies to:
\[ \alpha - \frac{\pi}{3} \]

5. **Combine the results**:
Now we can combine both parts:
\[ \cos^{-1}(x) + \cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3 - 3x^2}}{2}\right) = \alpha + \left(\alpha - \frac{\pi}{3}\right) = 2\alpha - \frac{\pi}{3} \]

6. **Substituting back for \( \alpha \)**:
Since \( \alpha = \cos^{-1}(x) \), we have:
\[ 2\cos^{-1}(x) - \frac{\pi}{3} \]

7. **Final Result**:
The final answer simplifies to:
\[ \frac{\pi}{3} \]