Q. If $\log_{\cos x}{\sin{x > 2\ and\ 0 < x < 3\pi,}}\$then $\sin x\$lies in the interval
A) $\left\lbrack \frac{\sqrt{5} - 1}{2},1 \right\rbrack$
B) $\left\lbrack 0,\frac{\sqrt{5} - 1}{2} \right\rbrack$
C) $\left\lbrack 0,\frac{1}{2} \right\rbrack$
D) None of these
Explanation
(b)
Since, $\sin{x \leq \cos^{2}{x,}becuase\cos x\ }$ must be a positive proper fraction
$\sin^{2}{x + \sin{x - 1 \leq 0}}$
Or $\left( \sin{x + \frac{1}{2}} \right)^{2} - \frac{5}{4} \leq 0$
From the definition of logarithm
$\sin{x > 0,\ \cos{x > 0,\ \cos{x \neq 1}}}$
$\therefore\sin{x + \frac{1}{2} \leq \frac{\sqrt{5}}{2},}$
$\Rightarrow 0 < \sin{x \leq \frac{\sqrt{5} - 1}{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.