If 2 tan ^2 theta-5 sec theta=1 has exactly 7 | vedclass

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(D) Given $2 	an ^2 	heta - 5 \sec 	heta = 1$.Using $	an ^2 	heta = \sec ^2 	heta - 1$,we get $2(\sec ^2 	heta - 1) - 5 \sec 	heta - 1 = 0$.$2

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$\frac{1}{2^{13}}(2^{14}-15)$

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(D) Given $2 	an ^2 	heta - 5 \sec 	heta = 1$.Using $	an ^2 	heta = \sec ^2 	heta - 1$,we get $2(\sec ^2 	heta - 1) - 5 \sec 	heta - 1 = 0$.$2 \sec ^2 	heta - 5 \sec 	heta - 3 = 0$.$(2 \sec 	heta + 1)(\sec 	heta - 3) = 0$.Since $\sec 	heta = -\frac{1}{2}$ is impossible,we have $\sec 	heta = 3$,which means $\cos 	heta = \frac{1}{3}$.In the interval $[0, 2\pi]$,there are $2$ solutions for $\cos 	heta = \frac{1}{3}$.For $7$ solutions,we need $3$ full periods of $2\pi$ plus one more solution in the next interval,so $n = 13$.We need to calculate $S = \sum_{k=1}^{13} \frac{k}{2^k}$.$S = \frac{1}{2} + \frac{2}{2^2} + \dots + \frac{13}{2^{13}}$.$\frac{1}{2}S = \frac{1}{2^2} + \dots + \frac{12}{2^{13}} + \frac{13}{2^{14}}$.Subtracting gives $\frac{S}{2} = \frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^{13}} - \frac{13}{2^{14}} = \frac{1/2(1 - 1/2^{13})}{1 - 1/2} - \frac{13}{2^{14}} = 1 - \frac{1}{2^{13}} - \frac{13}{2^{14}} = 1 - \frac{15}{2^{14}}$.$S = 2 - \frac{15}{2^{13}} = \frac{2^{14} - 15}{2^{13}}$.

If $2 \tan ^2 \theta-5 \sec \theta=1$ has exactly $7$ solutions in the interval $\left[0, \frac{n \pi}{2}\right]$,for the least value of $n \in N$,then $\sum_{k=1}^{n} \frac{k}{2^{k}}$ is equal to :

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