Number of solutions of equation |cot x|=cot x+frac{1}{sin x} in interval [0,2 π] is

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Trigonometrical Equations

hard

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

A

$1$

B

$2$

C

$3$

D

$4$

(JEE MAIN-2021)

Solution

If $\cot x>0 \Rightarrow \frac{1}{\sin x}=0$ (Not possible)

If $\operatorname{cotx}<0 \Rightarrow 2 \cot x+\frac{1}{\sin x}=0$

$\Rightarrow 2 \cos x=-1$

$\Rightarrow x =\frac{2 \pi}{3}$ or $\frac{4 \pi}{3}( reject )$

Standard 11

Mathematics

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