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(D) Let $S = \cos^{2} x + \cos^{4} x + \cos^{6} x + \dots \infty$. This is an infinite geometric series with first term $a = \cos^{2} x$ and common ra

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$\frac{1}{2}$

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(D) Let $S = \cos^{2} x + \cos^{4} x + \cos^{6} x + \dots \infty$. This is an infinite geometric series with first term $a = \cos^{2} x$ and common ratio $r = \cos^{2} x$.Since $0 The given expression is $e^{S \log_{e} 2} = e^{\log_{e} 2^{S}} = 2^{S} = 2^{\cot^{2} x}$.Given $t^{2} - 9t + 8 = 0$,we have $(t - 8)(t - 1) = 0$,so $t = 8$ or $t = 1$.Thus,$2^{\cot^{2} x} = 8 = 2^{3}$ or $2^{\cot^{2} x} = 1 = 2^{0}$.If $\cot^{2} x = 3$,then $\cot x = \sqrt{3}$ (since $x$ is in the first quadrant).If $\cot^{2} x = 0$,then $\cot x = 0$,which implies $x = \frac{\pi}{2}$,but $x So,$\cot x = \sqrt{3}$,which means $	an x = \frac{1}{\sqrt{3}}$,so $x = 30^{\circ}$ or $\frac{\pi}{6}$.Now,evaluate $\frac{2 \sin x}{\sin x + \sqrt{3} \cos x} = \frac{2}{\frac{\sin x}{\sin x} + \sqrt{3} \frac{\cos x}{\sin x}} = \frac{2}{1 + \sqrt{3} \cot x}$.Substituting $\cot x = \sqrt{3}$,we get $\frac{2}{1 + \sqrt{3}(\sqrt{3})} = \frac{2}{1 + 3} = \frac{2}{4} = \frac{1}{2}$.

If $e^{(\cos^{2} x + \cos^{4} x + \cos^{6} x + \dots \infty) \log_{e} 2}$ satisfies the equation $t^{2} - 9t + 8 = 0$,then the value of $\frac{2 \sin x}{\sin x + \sqrt{3} \cos x}$ for $0 < x < \frac{\pi}{2}$ is

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