A) x
B) \[-x\]
C) \[2\pi +x\]
D) \[2\pi -x\]
Correct Answer: D
Solution :
It is given that \[\pi \le x\le 2\pi \] \[\Rightarrow \,\,-\pi \ge -x\ge -2\pi \,\,\,\Rightarrow \,\,\pi \ge 2\pi -x\ge 0\] \[\Rightarrow \,\,{{\cos }^{-1}}\,\left\{ \cos \,(2\pi -x) \right\}=2\pi -x\].
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