If frac{1}{2} leq x leq 1, then cos ^{-1} x+c | vedclass

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(B) Let $x = \cos 	heta$. Since $\frac{1}{2} \leq x \leq 1$, we have $0 \leq 	heta \leq \frac{\pi}{3}$.Then, the expression becomes $\cos ^{-1}(\cos

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$\frac{\pi}{3}$

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(B) Let $x = \cos 	heta$. Since $\frac{1}{2} \leq x \leq 1$, we have $0 \leq 	heta \leq \frac{\pi}{3}$.Then, the expression becomes $\cos ^{-1}(\cos 	heta) + \cos ^{-1}\left(\frac{1}{2} \cos 	heta + \frac{\sqrt{3}}{2} \sin 	hetaight)$.Using the identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$, we set $\cos A = \frac{1}{2}$ and $\sin A = \frac{\sqrt{3}}{2}$, so $A = \frac{\pi}{3}$.Thus, the expression is $	heta + \cos ^{-1}(\cos(	heta - \frac{\pi}{3}))$.Since $0 \leq 	heta \leq \frac{\pi}{3}$, then $-\frac{\pi}{3} \leq 	heta - \frac{\pi}{3} \leq 0$, which implies $0 \leq \frac{\pi}{3} - 	heta \leq \frac{\pi}{3}$.Since $\cos(	heta - \frac{\pi}{3}) = \cos(\frac{\pi}{3} - 	heta)$, we have $\cos ^{-1}(\cos(	heta - \frac{\pi}{3})) = \frac{\pi}{3} - 	heta$.Therefore, the expression is $	heta + (\frac{\pi}{3} - 	heta) = \frac{\pi}{3}$.

If $\frac{1}{2} \leq x \leq 1$, then $\cos ^{-1} x+\cos ^{-1}\left(\frac{x}{2}+\frac{1}{2} \sqrt{3-3 x^2}\right)$ is equal to

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