Question

Find the number of solutions of the equations;

$\left|\cot x\right|=\cot x+\frac{1}{\sin x}$ ,where $x\in [0,2\pi ]$

Answer 1

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

For $\cot x\ge 0\Rightarrow x\in (n\pi ,n\pi \frac{\pi }{2})$

$\cot x=\cot x+\frac{1}{\sin x}$$\Rightarrow cscx=0$

And for $\cot x<0\Rightarrow xϵ(n\pi \frac{-\pi }{2},n\pi )$

$-\cot x=\cot x\frac{+1}{\sin x}\Rightarrow 2\cot x\frac{+1}{\sin x}=0$

$\Rightarrow \frac{2cosx}{\sin x}+\frac{1}{\sin x}=0\Rightarrow cosx=\frac{-1}{2}$

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

Hence, there are three solutions in $[0,3\pi ]$