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  • The number of solutions of the equation <img alt="|\cot x|=\cot x+\frac{1}{\sin x}(0 \leq x \leq 2 \pi)" src="https://entrancecorner.oncodecogs.com/gif.latex?%7C%5Ccot%20x%7C%3D%5Ccot%20x

The number of solutions of the equation |\cot x|=\cot x+\frac{1}{\sin x}(0 \leq x \leq 2 \pi) is

Option: 1

0
 


Option: 2

1


Option: 3

2

 


Option: 4

3


Answers (1)

best_answer

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

If x lies in I and III quadrant, then

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

\Rightarrow \quad \frac{1}{\sin x}=0, not possible

and, if  x lies in II and IV quadrant, then
-\cot x=\cot x+\frac{1}{\sin x}
\text { or } \quad 2 \cos x+1=0, \quad \because \sin x \neq 0
\cos x=-\frac{1}{2}=\cos \frac{2 \pi}{3}

\text { or } \quad x=2 n \pi \pm \frac{2 \pi}{3}
\text { Then, } \quad x=\frac{2 \pi}{3}, \frac{4 \pi}{3} \quad(\because x \in[0,2 \pi])
Hence, no. of solutions =2

 

 

 

 

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Gunjita

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