Question:
The number of solutions of the equation $|\cot x|=\cot x+\frac{1}{\sin x}$ in the interval $[0,2 \pi]$ is
Solution:
If $\cot x>0 \Rightarrow \frac{1}{\sin x}=0$ (Not possible)
If $\cot x<0 \Rightarrow 2 \cot x+\frac{1}{\sin x}=0$
$\Rightarrow 2 \cos x=-1$
$\Rightarrow \mathrm{x}=\frac{2 \pi}{3}$ or $\frac{4 \pi}{3}$ (reject)
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