Let S = {theta in (0, 2pi) : 7 cos^2 theta - | vedclass

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(D) Given the equation: $7 \cos^2 	heta - 3 \sin^2 	heta - 2 \cos^2 2	heta = 2$.Using $\sin^2 	heta = 1 - \cos^2 	heta$,we get $7 \cos^2 	heta -

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$16$

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(D) Given the equation: $7 \cos^2 	heta - 3 \sin^2 	heta - 2 \cos^2 2	heta = 2$.Using $\sin^2 	heta = 1 - \cos^2 	heta$,we get $7 \cos^2 	heta - 3(1 - \cos^2 	heta) - 2 \cos^2 2	heta = 2$,which simplifies to $10 \cos^2 	heta - 3 - 2 \cos^2 2	heta = 2$.Since $2 \cos^2 	heta = 1 + \cos 2	heta$,we have $5(1 + \cos 2	heta) - 3 - 2 \cos^2 2	heta = 2$,leading to $5 + 5 \cos 2	heta - 3 - 2 \cos^2 2	heta = 2$.This simplifies to $2 \cos^2 2	heta - 5 \cos 2	heta = 0$,so $\cos 2	heta(2 \cos 2	heta - 5) = 0$.Since $\cos 2	heta$ cannot be $2.5$,we have $\cos 2	heta = 0$,which gives $2	heta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$.Thus,$S = \{\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\}$.For any $	heta \in S$,$	an^2 	heta = 1$ and $\cot^2 	heta = 1$,so $	an^2 	heta + \cot^2 	heta = 2$.Also,$\sin^2 	heta = \frac{1}{2}$.The equation becomes $x^2 - 2(2)x + 6(\frac{1}{2}) = 0$,which is $x^2 - 4x + 3 = 0$.The sum of roots for each equation is $4$.Since there are $4$ such equations,the total sum of roots is $4 	imes 4 = 16$.

Let $S = \{\theta \in (0, 2\pi) : 7 \cos^2 \theta - 3 \sin^2 \theta - 2 \cos^2 2\theta = 2\}$. Then,the sum of roots of all the equations $x^2 - 2(\tan^2 \theta + \cot^2 \theta)x + 6 \sin^2 \theta = 0$ for $\theta \in S$ is...

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