Number of values of x in [0, 2pi] satisfying | vedclass

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(C) Given equation: $\cot x - \cos x = 1 - \cot x \cos x$Rearranging the terms: $\cot x - 1 - \cos x + \cot x \cos x = 0$$\cot x(1 + \cos x) - 1(1 + \

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$2$

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(C) Given equation: $\cot x - \cos x = 1 - \cot x \cos x$Rearranging the terms: $\cot x - 1 - \cos x + \cot x \cos x = 0$$\cot x(1 + \cos x) - 1(1 + \cos x) = 0$$(\cot x - 1)(1 + \cos x) = 0$This implies $\cot x = 1$ or $\cos x = -1$.Case $1$: $\cot x = 1 \implies 	an x = 1$. In the interval $[0, 2\pi]$,$x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$.Case $2$: $\cos x = -1$. In the interval $[0, 2\pi]$,$x = \pi$.Note that at $x = \pi$,$\cot x$ is defined as $\frac{\cos \pi}{\sin \pi}$,which is undefined. Thus,$x = \pi$ is not a solution.The valid solutions are $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$.Therefore,the number of values is $2$.

Number of values of $x \in [0, 2\pi]$ satisfying the equation $\cot x - \cos x = 1 - \cot x \cos x$.

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