Question

Total number of solution of $\left|\cot x\right|=\cot x+\frac{1}{\sin x}\times \in \left[0,3\pi \right],$ is :-

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

The correct option is C $3$

$\left|\cot x\right|=\cot x+\frac{1}{\sin x}$

Case$1$

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

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

$\sin x=\infty$ is not possible

Case$2$

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

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

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

$\Rightarrow \cos x=-\frac{1}{2}$ lies in second and third quadrant.

And $x\in [0,3\pi ]$

$\therefore x=\frac{\pi }{3},\pi +\frac{\pi }{3},3\pi -\frac{\pi }{3}$

or $x=\{\frac{\pi }{3},\frac{4\pi }{3},\frac{8\pi }{3}\}$

$\therefore$ there are $3$ solutions.