Let X = {x in mathbb{R} : cos(sin x) = sin(co | vedclass

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(A) Given the equation $\cos(\sin x) = \sin(\cos x)$.We know that $\sin(	heta) = \cos(\frac{\pi}{2} - 	heta)$.So,$\cos(\sin x) = \cos(\frac{\pi}{2}

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(A) Given the equation $\cos(\sin x) = \sin(\cos x)$.We know that $\sin(	heta) = \cos(\frac{\pi}{2} - 	heta)$.So,$\cos(\sin x) = \cos(\frac{\pi}{2} - \cos x)$.This implies $\sin x = 2n\pi \pm (\frac{\pi}{2} - \cos x)$ for $n \in \mathbb{Z}$.Case $1$: $\sin x + \cos x = 2n\pi + \frac{\pi}{2}$.The range of $\sin x + \cos x$ is $[-\sqrt{2}, \sqrt{2}] \approx [-1.414, 1.414]$.For $n=0$,$\frac{\pi}{2} \approx 1.57$,which is outside the range.Case $2$: $\sin x - \cos x = 2n\pi - \frac{\pi}{2}$.The range of $\sin x - \cos x$ is $[-\sqrt{2}, \sqrt{2}] \approx [-1.414, 1.414]$.For $n=0$,$-\frac{\pi}{2} \approx -1.57$,which is outside the range.Since no value of $n$ satisfies the equation,the number of elements in $X$ is $0$.

Let $X = \{x \in \mathbb{R} : \cos(\sin x) = \sin(\cos x)\}$. The number of elements in $X$ is

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