Question

If ${\log }_{\cos x}\sin \ge 2$ and $0\le x\le 3\pi$, then $\sin x$ lies in the interval

• A. $[\frac{\sqrt{5}-1}{2},1]$
• B. $[0,\frac{\sqrt{5}-1}{2}]$
• C. $[0,\frac{1}{2}]$
• D. none of these

Answer 1

The correct option is B $[0,\frac{\sqrt{5}-1}{2}]$

$\sin x\ge {\cos }^{2}x\sin x>0\sin x\ge 1-{\sin }^{2}x{\sin }^{2}x+\sin x-1\ge 0[\sin x=\frac{-1\pm \sqrt{5}}{2}][\sin x-\left(\frac{-1+\sqrt{5}}{2}\right)][\sin x-\left(\frac{-1-\sqrt{5}}{2}\right)]\ge 0$

But $\sin x$ should be positive

Hence, $\sin x\in [0,\frac{\sqrt{5}-1}{2}]$