Question

If $lo{g}_{cosx}sinx\ge 2$ and $x\in [0,3\pi ]$, then $sinx$ lies in interval

• A. $[\frac{\sqrt{5}-1}{2},1]$
• B. $(0,\frac{\sqrt{5}-1}{2})$
• C. $[0,\frac{1}{2}]$
• D. $(-1,0)$

Answer 1

Answer: (A)

$[\frac{\sqrt{5}-1}{2},1]$

$\frac{\log \sin x}{\log \cos x}\ge 2$

$\Rightarrow$$\log \sin x\ge 2\log \cos x$

$\Rightarrow$$\sin x\ge e^{\log {\cos }^{2}x}$

$\Rightarrow$$\sin x\ge {\cos }^{2}x$

$\Rightarrow$${\sin }^{2}x+\sin x-1\ge 0$

Hence $\sin x=\frac{-1\pm \sqrt{1-4(-1)\left(1\right)}}{2}$

Hence expression reduces to

$(\sin x-\frac{\sqrt{5}-1}{2})(\sin x+\frac{\sqrt{5}+1}{2})\ge 0$

Either both brackets should be positive or both negative but 2nd bracket can never be negative as range of $\sin x$ is $[-1,1]$

Hence considering both need to be positve and maximum value of $\sin x$ is only 1 option A gives correct range of values