Question

The total number of solutions of $\left|\cot x\right|=\cot x+\frac{1}{\sin x},xϵ[0,3\pi ]$, is equal to

• A. 1
• B. 2
• C. 3
• D. 0

Answer 1

The correct option is B 2

$|\cot x|=\cot x+\frac{1}{\sin x},xϵ[0,3\pi ]$
If $x\in [0,\frac{\pi }{2}]\cup [\pi ,\frac{3\pi }{2}]\cup [2\pi ,\frac{5\pi }{2}]$
$\Rightarrow |\cot x|=\cot x$
So, $\cot x=\cot x+\frac{1}{\sin x}\Rightarrow \frac{1}{\sin x}=0\Rightarrow \sin x=\pm \infty$
but we know $-1\le \sin x\le 1$, thus there is no solution in this interval
If $x\in (\frac{\pi }{2},\pi )\cup (\frac{3\pi }{2},2\pi )\cup (\frac{5\pi }{2},3\pi ]$
$\Rightarrow \mid \cot x\mid =-\cot x$
So, $-\cot x=\cot x+\frac{1}{\sin x}\Rightarrow \frac{1}{\sin x}=-2\cot x\Rightarrow 2\cot x\sin x+1=0$

$\Rightarrow 2\cos x+1=0$
$\Rightarrow \cos x=-\frac{1}{2}$
Therefore solutions in the given interval are,
$x=\frac{2\pi }{3},\frac{4\pi }{3}$
Hence, option 'B' is correct.