The number of solutions of |cos x| = sin x in | vedclass

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(C) We need to solve $|\cos x| = \sin x$ for $x \in [-4 \pi, 4 \pi]$.Since $|\cos x| \geq 0$,we must have $\sin x \geq 0$. This implies $x$ must be in

Vedclass · Accepted Answer

$8$

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(C) We need to solve $|\cos x| = \sin x$ for $x \in [-4 \pi, 4 \pi]$.Since $|\cos x| \geq 0$,we must have $\sin x \geq 0$. This implies $x$ must be in the first or second quadrant.Case $1$: $\cos x = \sin x \implies 	an x = 1$. In $[0, 2 \pi]$,this gives $x = \frac{\pi}{4}, \frac{5 \pi}{4}$. Since $\sin x$ must be positive,only $x = \frac{\pi}{4}$ is a solution.Case $2$: $-\cos x = \sin x \implies 	an x = -1$. In $[0, 2 \pi]$,this gives $x = \frac{3 \pi}{4}, \frac{7 \pi}{4}$. Since $\sin x$ must be positive,only $x = \frac{3 \pi}{4}$ is a solution.Thus,there are $2$ solutions in every interval of length $2 \pi$ (specifically $[0, 2 \pi]$ and $[-2 \pi, 0]$).In the interval $[-4 \pi, 4 \pi]$,which consists of $4$ intervals of length $2 \pi$,the total number of solutions is $2 	imes 4 = 8$.

The number of solutions of $|\cos x| = \sin x$ in the interval $-4 \pi \leq x \leq 4 \pi$ is:

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