Question

The number of solutions of the equation $\left|\cot \left(x\right)\right|=\cot \left(x\right)+\frac{1}{\sin \left(x\right)}$ in the interval $\left[0,2\pi \right]$ is

Answer 1

Find the number of solution

Case (i): $x\in \left[0,\frac{\pi }{2}\right]\cup \left[\pi ,\frac{3\pi }{2}\right]$

$\cot \left(x\right)=\cot \left(x\right)+\frac{1}{\sin \left(x\right)}$, which is not possible

Case (ii): $x\in \left[\frac{\pi }{2},\pi \right]\cup \left[\frac{3\pi }{2},2\pi \right]$

$$\begin{gathered} -\cot \left(x\right)=\cot \left(x\right)+\frac{1}{\sin \left(x\right)} \\ \Rightarrow -2\cot \left(x\right)=\frac{1}{\sin \left(x\right)} \\ \Rightarrow -2\frac{\cos \left(x\right)}{\sin \left(x\right)}=\frac{1}{\sin \left(x\right)} \\ \Rightarrow \cos \left(x\right)=-\frac{1}{2} \\ \therefore x=\frac{2\pi }{3},\frac{4\pi }{3} \end{gathered}$$

Hence, number of solution of the equation is $2$.