A) \[\left[ 0,\frac{\pi }{6} \right]\]
B) \[\left[ 0,\frac{5\pi }{6} \right]\]
C) \[\left[ \frac{5\pi }{6},2\pi \right]\]
D) \[\left[ \frac{5\pi }{6},\frac{\pi }{6} \right]\]
Correct Answer: D
Solution :
We have, \[4{{\sin }^{2}}x-8\sin x+3\le 0,0\le x\le 2\,\pi \] \[\Rightarrow \] \[(2\sin x-1)(2\sin x-3)\le 0\] But \[2\sin x-3\]is always negative, as \[\sin \,x\le 1\] \[\therefore \] \[2\sin x-1\ge 0\] \[\Rightarrow \] \[\sin x\ge \frac{1}{2}\] \[\therefore \] \[\frac{\pi }{6}\le x\le \frac{5\pi }{6}\] Hence, \[x\in \left[ \frac{\pi }{6},\frac{5\pi }{6} \right]\]
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