The number of solutions of the equation |cot | vedclass

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(A) The given equation is $|\cot x| = \cot x + \frac{1}{\sin x}$.Case $1$: If $\cot x \ge 0$,then $|\cot x| = \cot x$. The equation becomes $\cot x =

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$1$

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(A) The given equation is $|\cot x| = \cot x + \frac{1}{\sin x}$.Case $1$: If $\cot x \ge 0$,then $|\cot x| = \cot x$. The equation becomes $\cot x = \cot x + \frac{1}{\sin x}$,which implies $\frac{1}{\sin x} = 0$. This has no solution.Case $2$: If $\cot x Substituting $\cot x = \frac{\cos x}{\sin x}$,we get $2\frac{\cos x}{\sin x} + \frac{1}{\sin x} = 0$,so $\frac{2\cos x + 1}{\sin x} = 0$.This implies $2\cos x + 1 = 0$,so $\cos x = -\frac{1}{2}$.In the interval $[0, 2\pi]$,$\cos x = -\frac{1}{2}$ at $x = \frac{2\pi}{3}$ and $x = \frac{4\pi}{3}$.Check the condition $\cot x For $x = \frac{4\pi}{3}$,$\cot(\frac{4\pi}{3}) = \frac{1}{\sqrt{3}} > 0$ (Invalid,as it contradicts $\cot x Thus,there is only $1$ solution.

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

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