What is the number of roots of the quadratic | vedclass

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(D) Given the equation $8\sec^2	heta - 6\sec	heta + 1 = 0$. Let $x = \sec	heta$. The equation becomes $8x^2 - 6x + 1 = 0$. Factoring the quadratic:

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$0$

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(D) Given the equation $8\sec^2	heta - 6\sec	heta + 1 = 0$. Let $x = \sec	heta$. The equation becomes $8x^2 - 6x + 1 = 0$. Factoring the quadratic: $8x^2 - 4x - 2x + 1 = 0 \implies 4x(2x - 1) - 1(2x - 1) = 0$. $(4x - 1)(2x - 1) = 0$. So,$x = \frac{1}{4}$ or $x = \frac{1}{2}$. This means $\sec	heta = \frac{1}{4}$ or $\sec	heta = \frac{1}{2}$. Since the range of $\sec	heta$ is $(-\infty, -1] \cup [1, \infty)$,the values $\frac{1}{4}$ and $\frac{1}{2}$ are not possible. Therefore,there are no real roots for $	heta$. The number of roots is $0$.

$List-I$	$List-II$
$(I)$ $\{x \in[-\frac{2 \pi}{3}, \frac{2 \pi}{3}]: \cos x+\sin x=1\}$	$(P)$ has two elements
$(II)$ $\{x \in[-\frac{5 \pi}{18}, \frac{5 \pi}{18}]: \sqrt{3} \tan 3 x=1\}$	$(Q)$ has three elements
$(III)$ $\{x \in[-\frac{6 \pi}{5}, \frac{6 \pi}{5}]: 2 \cos (2 x)=\sqrt{3}\}$	$(R)$ has four elements
$(IV)$ $\{x \in[-\frac{7 \pi}{4}, \frac{7 \pi}{4}]: \sin x-\cos x=1\}$	$(S)$ has five elements
	$(T)$ has six elements

What is the number of roots of the quadratic equation $8\sec^2\theta - 6\sec\theta + 1 = 0$?

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