Let x in [0, 6pi] satisfy the equation cos x | vedclass

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(A) Given equation: $\cos x - \sin x = -1$.Divide by $\sqrt{2}$: $\frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x = -\frac{1}{\sqrt{2}}$.Using $

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$3$

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(A) Given equation: $\cos x - \sin x = -1$.Divide by $\sqrt{2}$: $\frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x = -\frac{1}{\sqrt{2}}$.Using $\cos(x + \frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$, we have $x + \frac{\pi}{4} = 2n\pi \pm \frac{3\pi}{4}$.Case $1$: $x + \frac{\pi}{4} = 2n\pi + \frac{3\pi}{4} \implies x = 2n\pi + \frac{\pi}{2}$.For $x \in [0, 6\pi]$, $n=0, 1, 2$ gives $x = \frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2}$.Case $2$: $x + \frac{\pi}{4} = 2n\pi - \frac{3\pi}{4} \implies x = 2n\pi - \pi$.For $x \in [0, 6\pi]$, $n=1, 2, 3$ gives $x = \pi, 3\pi, 5\pi$.Given $x = k(\frac{\pi}{3})$, so $k = \frac{3x}{\pi}$.For $x = \frac{\pi}{2}, k = 1.5$ (not in $\mathbb{N}$).For $x = \frac{5\pi}{2}, k = 7.5$ (not in $\mathbb{N}$).For $x = \frac{9\pi}{2}, k = 13.5$ (not in $\mathbb{N}$).For $x = \pi, k = 3$.For $x = 3\pi, k = 9$.For $x = 5\pi, k = 15$.Thus, $k \in \{3, 9, 15\}$. The number of values is $3$.

Let $x \in [0, 6\pi]$ satisfy the equation $\cos x - \sin x = -1$. If $x = k(\frac{\pi}{3})$ where $k \in \mathbb{N}$, then find the number of possible values of $k$.

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