Question

The number of solutions of the equation $\left|cotx\right|=cotx+\frac{1}{sinx}$ in $0\le x\le 2\pi ,$ is :-

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

If $cotx>0,$ then $\frac{1}{sinx}=0$ which is not possible
If $cotx<0,$ then we have
$-cotx=cotx+\frac{1}{sinx}$
$\Rightarrow 0=\frac{2cosx}{sinx}+\frac{1}{sinx}$
$\Rightarrow \frac{2cosx+1}{sinx}=0$
$\Rightarrow cosx=-\frac{1}{2}=cos\frac{2\pi }{3}\&cotx<0$
So, $x$ lies in $I{I}^{nd}$ quadrants
$\therefore x=2n\pi \pm \frac{2\pi }{3},n\in I$
and $\because 0\le x\le 2\pi \therefore x=\frac{2\pi }{3}$
$\therefore$ Number of solutions $=1$