Question

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

• A. $1$
• B. $3$
• C. $2$
• D. $8$

Answer 1

The correct option is C $2$

${\log }_{0.5}=1-{\log }_{0.5}\cos x$

$\Rightarrow {\log }_{0.5}\left(\sin x\right)+{\log }_{0.5}\left(\cos x\right)=1$

$\Rightarrow {\log }_{\left(0.5\right)}\left(\sin x\cos x\right)=1$

$\Rightarrow {\left(\frac{1}{2}\right)}^1=\sin x\cos x$

$\Rightarrow \frac{1}{2}=\frac{1}{2}\times 2\sin x\cos x$

$\Rightarrow \sin 2x=1$

$\Rightarrow 2x\in (2x+1)\frac{\pi }{2}$

$\Rightarrow x\in \left[\right(2n+1\left)\frac{\pi }{4}\right]$

No.we chat the given ray

$-2\pi \le x\le 2\pi$

$\therefore x\in [\frac{-7\pi }{4},\frac{-5\pi }{4},\frac{-3\pi }{4},\frac{-\pi }{4}]$

$\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$

Also, ${\log }_{0.5}\left(\sin x\right)>0$

${\log }_{0.5}\left(\cos x\right)>0$

$\therefore xt[\frac{-7\pi }{4},\frac{\pi }{4}]$

$\therefore$ there are $2$ solution in the given rays