Let tan ^{-1}(x) inleft(-frac{pi}{2}, frac{pi | vedclass

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(D) Given equation: $\sqrt{1+\cos (2 x)}=\sqrt{2} 	an ^{-1}(	an x)$Using the identity $1+\cos(2x) = 2\cos^2 x$,we get $\sqrt{2\cos^2 x} = \sqrt{2} \

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$3$

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(D) Given equation: $\sqrt{1+\cos (2 x)}=\sqrt{2} 	an ^{-1}(	an x)$Using the identity $1+\cos(2x) = 2\cos^2 x$,we get $\sqrt{2\cos^2 x} = \sqrt{2} 	an^{-1}(	an x)$.This simplifies to $\sqrt{2}|\cos x| = \sqrt{2} 	an^{-1}(	an x)$,or $|\cos x| = 	an^{-1}(	an x)$.We need to find the number of intersection points of $y = |\cos x|$ and $y = 	an^{-1}(	an x)$ in the domain $D = \left(-\frac{3 \pi}{2},-\frac{\pi}{2}ight) \cup\left(-\frac{\pi}{2}, \frac{\pi}{2}ight) \cup\left(\frac{\pi}{2}, \frac{3 \pi}{2}ight)$.$1$. In $\left(-\frac{\pi}{2}, \frac{\pi}{2}ight)$,$	an^{-1}(	an x) = x$. The equation is $|\cos x| = x$. Since $|\cos x| > 0$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}ight) \setminus \{0\}$ and $x=0$ gives $|\cos 0| = 1 
eq 0$,there is one solution for $x > 0$ and no solution for $x \leq 0$.$2$. In $\left(\frac{\pi}{2}, \frac{3 \pi}{2}ight)$,$	an^{-1}(	an x) = x - \pi$. The equation is $|\cos x| = x - \pi$. There is one intersection point in this interval.$3$. In $\left(-\frac{3 \pi}{2}, -\frac{\pi}{2}ight)$,$	an^{-1}(	an x) = x + \pi$. The equation is $|\cos x| = x + \pi$. There is one intersection point in this interval.Total number of solutions is $1 + 1 + 1 = 3$.

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