If 2 tan^2 theta - 4 sec theta + 3 = 0,then 2 | vedclass

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(D) Given the equation $2 	an^2 	heta - 4 \sec 	heta + 3 = 0$. Using the identity $	an^2 	heta = \sec^2 	heta - 1$,we substitute: $2(\sec^2 	he

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$2 + \sqrt{2}$

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(D) Given the equation $2 	an^2 	heta - 4 \sec 	heta + 3 = 0$. Using the identity $	an^2 	heta = \sec^2 	heta - 1$,we substitute: $2(\sec^2 	heta - 1) - 4 \sec 	heta + 3 = 0$ $2 \sec^2 	heta - 2 - 4 \sec 	heta + 3 = 0$ $2 \sec^2 	heta - 4 \sec 	heta + 1 = 0$. Let $x = \sec 	heta$. Then $2x^2 - 4x + 1 = 0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{4 \pm \sqrt{16 - 8}}{4} = \frac{4 \pm \sqrt{8}}{4} = \frac{4 \pm 2\sqrt{2}}{4} = 1 \pm \frac{\sqrt{2}}{2}$. Thus,$\sec 	heta = 1 + \frac{\sqrt{2}}{2}$ or $\sec 	heta = 1 - \frac{\sqrt{2}}{2}$. Since $|\sec 	heta| \ge 1$,we must have $\sec 	heta = 1 + \frac{\sqrt{2}}{2} = \frac{2 + \sqrt{2}}{2}$. Therefore,$2 \sec 	heta = 2 + \sqrt{2}$.

If $2 \tan^2 \theta - 4 \sec \theta + 3 = 0$,then $2 \sec \theta =$

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