Question

If ${\log }_{\cos x}\sin x\ge 2$ and $x\in (0,3\pi )-\frac{n\pi }{2},n\in I$, 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 $(0,\frac{\sqrt{5}-1}{2}]$

${\log }_{\cos x}\sin x\ge 2$
$\Rightarrow \sin x\le {\cos }^{2}x$, since base of log is less than 1.
$\Rightarrow \sin x\le 1-{\sin }^{2}x$
$\Rightarrow {\sin }^{2}x+\sin x-1\le 0$
$\Rightarrow \sin x\in [-\frac{\sqrt{5}+1}{2},\frac{\sqrt{5}-1}{2}]$
Also check the domain $\sin x>0$ and $\cos x>0,\ne 1$
Thus solution is, $\sin x\in (0,\frac{\sqrt{5}-1}{2}]$