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: $\cot x - \cos x + \cot x \cos x - 1 = 0$.Group the terms: $\cot x(1

Vedclass · Accepted Answer

$3$

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(C) Given equation: $\cot x - \cos x = 1 - \cot x \cos x$.Rearranging the terms: $\cot x - \cos x + \cot x \cos x - 1 = 0$.Group the terms: $\cot x(1 + \cos x) - (1 + \cos x) = 0$.Factor out $(1 + \cos x)$: $(1 + \cos x)(\cot x - 1) = 0$.This implies either $1 + \cos x = 0$ or $\cot x - 1 = 0$.Case $1$: $\cos x = -1$. In the interval $[0, 2\pi]$,$x = \pi$.Case $2$: $\cot x = 1$. In the interval $[0, 2\pi]$,$x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$.The values of $x$ are $\frac{\pi}{4}, \pi, \frac{5\pi}{4}$.Thus,there are $3$ values of $x$ that satisfy the equation.

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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