The value of $\cos^{-1} x + \cos^{-1}\left(\frac{x}{2} + \frac{1}{2}\sqrt{3-3x^2}\right)$ is equal to: $\left(\frac{1}{2} \leq x \leq 1\right)$
Step-by-Step Solution
Key Concept: Inverse trigonometric identities and algebraic manipulation connect domain constraints to specific angle values.
Given $\cos^{-1}x = y$ with $\frac{1}{2} \leq x \leq 1$, we have $0 \leq y \leq \frac{\pi}{3}$. From $x = \cos y$, derive $\sin y = \sqrt{1-x^2}$. Using the constraint $\frac{x}{2} + \frac{1}{2}\sqrt{3-3x^2} = \frac{\sqrt{3}}{2}\sqrt{1-x^2}$, solve to find $\cos^{-1}x + \cos^{-1}(\frac{1}{2} - \frac{1}{2}\sqrt{3-3x^2}) = y + \frac{\pi}{3} - y = \frac{\pi}{3}$.
Correct Answer: 2