Question

If ${\log }_{0.5}\sin x=1-{\log }_{0.5}\cos x$, then the number of solutions of $x\in [-2\pi ,2\pi ]$ is/are

• A. $3$
• B. $2$
• C. $1$
• D. $4$

Answer 1

The correct option is B $2$

${\log }_{0.5}\sin x=1-{\log }_{0.5}\cos x$ where $x\in [-2\pi ,2\pi ]$
Above equation is valid when, $\sin x>0,$
$\Rightarrow x\in (-2\pi ,-\pi )\cup (0,\pi )$
and $\cos x>0$
$\Rightarrow x\in (-2\pi ,-\frac{3\pi }{2})\cup (-\frac{\pi }{2},\frac{\pi }{2})\cup (\frac{3\pi }{2},2\pi )$
Therefore, $x\in (-2\pi ,-\frac{3\pi }{2})\cup (0,\frac{\pi }{2})$
${\log }_{0.5}\sin x=1-{\log }_{0.5}\cos x$
$\Rightarrow {\log }_{0.5}\sin x+{\log }_{0.5}\cos x=1$
$\Rightarrow {\log }_{0.5}\left(\sin x\cos x\right)=1,[\because \log a+\log b=\log (ab\left)\right]$
$\Rightarrow \sin x\cos x={0.5}^1=\frac{1}{2}$
$\Rightarrow 2\sin x\cos x=1$
$\Rightarrow \sin 2x=1$
$\Rightarrow \cos 2x=0$
$\Rightarrow 1-2{\sin }^{2}x=0$
$\Rightarrow {\sin }^{2}x=\frac{1}{2}$
$\Rightarrow \sin x=\pm \frac{1}{\sqrt{2}}$
Now, $\because x\in (-2\pi ,-\frac{3\pi }{2})\cup (0,\frac{\pi }{2})$
$\therefore x=\{-\frac{7\pi }{4},\frac{\pi }{4}\}$
Hence there are two solutions in the given interval
Ans: B