Question

The total no. of solution of equation
$|cotx|=cotx+\frac{1}{sinx},x\in [0,3\pi ]$ is equal to

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is A $3$

Given, $\left|\cot x\right|=\cot x+\frac{1}{\sin x}$
For $\cot x\ge 0\Rightarrow xϵ(n\pi ,n\pi \frac{\pi }{2})$
$\cot x=\cot x+\frac{1}{\sin x}$
$\Rightarrow cscx=0$
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 in $[0,3\pi ]$ 3 solutions