Let [x] denote the largest integer leq x. If | vedclass

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(C) Given equation: $\sin x \sqrt{4 \cos^2 x} = \frac{2+x-[x]}{1-x+[x]}$.Since $x-[x] = \{x\}$,the equation becomes $\sin x \cdot 2|\cos x| = \frac{2+

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is equal to zero

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(C) Given equation: $\sin x \sqrt{4 \cos^2 x} = \frac{2+x-[x]}{1-x+[x]}$.Since $x-[x] = \{x\}$,the equation becomes $\sin x \cdot 2|\cos x| = \frac{2+\{x\}}{1-\{x\}}$.For $x \in \left[\frac{\pi}{4}, \frac{\pi}{3}ight]$,$\cos x > 0$,so $|\cos x| = \cos x$.The equation simplifies to $\sin 2x = \frac{2+\{x\}}{1-\{x\}}$.In the interval $x \in \left[\frac{\pi}{4}, \frac{\pi}{3}ight]$,the maximum value of $\sin 2x$ is $\sin\left(\frac{2\pi}{3}ight) = \frac{\sqrt{3}}{2} \approx 0.866$.The expression $f(\{x\}) = \frac{2+\{x\}}{1-\{x\}}$ for $\{x\} \in [0, 1)$ has a minimum value at $\{x\} = 0$,which is $f(0) = \frac{2+0}{1-0} = 2$.Since the maximum value of the $LHS$ $(0.866)$ is less than the minimum value of the $RHS$ $(2)$,there are no solutions for $x$ in the given interval.Thus,$k = 0$.Therefore,$k^{	an^2 x} = 0^{	an^2 x} = 0$ for all $x \in \left[\frac{\pi}{4}, \frac{\pi}{3}ight]$.

Let $[x]$ denote the largest integer $\leq x$. If the number of solutions of $\sin x \sqrt{4 \cos ^2 x} = \frac{2+x-[x]}{1-x+[x]}$ is $k$,then for $x \in \left[\frac{\pi}{4}, \frac{\pi}{3}\right]$,the value of $k^{\tan^2 x}$

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