If pi leq x leq 2 pi,then cos^{-1}(cos x) is | vedclass

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(D) We know that the range of the principal value branch of $\cos^{-1} x$ is $[0, \pi]$.Given the interval $\pi \leq x \leq 2 \pi$.We use the property

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$2 \pi - x$

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(D) We know that the range of the principal value branch of $\cos^{-1} x$ is $[0, \pi]$.Given the interval $\pi \leq x \leq 2 \pi$.We use the property $\cos(2 \pi - x) = \cos x$.Since $\pi \leq x \leq 2 \pi$,we have $-\pi \geq -x \geq -2 \pi$.Adding $2 \pi$ to all sides,we get $2 \pi - \pi \geq 2 \pi - x \geq 2 \pi - 2 \pi$,which simplifies to $\pi \geq 2 \pi - x \geq 0$.Thus,$0 \leq 2 \pi - x \leq \pi$,which lies within the principal range $[0, \pi]$.Therefore,$\cos^{-1}(\cos x) = \cos^{-1}(\cos(2 \pi - x)) = 2 \pi - x$.

If $\pi \leq x \leq 2 \pi$,then $\cos^{-1}(\cos x)$ is equal to :-

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