Question

If $ 4{{\sin }^{2}}x-8\sin x+3=0, $ $ 0\le x\le 2\pi , $ then the solution set for $ x $ is

• 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] $

Answer 1

Answer: (D)

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] $