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

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

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$\frac{\pi}{3}$

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