Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

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