Question

If $n$ be the number of solutions of the equation $|\cot x|=\cot x+\frac{1}{\sin x}(0<x<2\pi )$ , then n $=$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

$\mid \cot x\mid =\cot x+\frac{1}{\sin x}0<x<2\pi$

$\mid \cot x\mid$ is positive if $\cot x>0$

Then,

$\cot x=\cot x+\frac{1}{\sin x}$

or, $\frac{1}{\sin x}=0$

or, $\sin x=\infty$ which is impossible.

$\mid \cot x\mid$ is -ve if $\cot x<0$

$-\cot x=\cot x+\frac{1}{\sin x}$

or,$-2\cot x=\frac{1}{\sin x}$

or, $\frac{-2\cos x}{\sin x}=\frac{1}{\sin x}$

$\sin x\ne 0$ because $x\ne 0$

or, $-2\cos x=1$

or, $\cos x=\frac{-1}{2}=\cos \frac{2\pi }{3}$

or, $x=2n\pi \pm \frac{2\pi }{3}$

If $n=1$

$x=2\pi -\frac{2\pi }{3}=\frac{4\pi }{3}$

which is in the range $(0,2\pi )$

$\therefore x=\frac{2\pi }{3},\frac{4\pi }{3}$

$\therefore n=2$