Let S = \ θ ∈ ( 0, 2π ) : 7 ^ 2 θ - 3 ^ 2 θ - 2 ^ 2 2θ = 2\ . Then,...

Given,

$7\cos^{2}\theta - 3\sin^{2}\theta - 2\cos^{2}2\theta = 2$

$⇒ 7\cos^{2}\theta - 3\left(1 - \cos^{2}\theta\right) - 2\cos^{2}2\theta = 2$

$⇒ 4\cos^{2}\theta + 3\cos 2\theta - 2\cos^{2}2\theta = 2$

$⇒ 2\left(1 + \cos 2\theta\right) + 3\cos 2\theta - 2\cos^{2}2\theta = 2$

$⇒ 2\cos^{2}2\theta - 5\cos 2\theta = 0$

$⇒ \cos 2\theta\left(2\cos 2\theta - 5\right) = 0$

$⇒ \cos 2\theta = 0$ or $2\cos 2\theta - 5 = 0\left\{\text{which is not possible}\right\}$

$⇒ 2\theta = \left(2n + 1\right)\dfrac{\pi}{2}$

$⇒ \theta = \left(2n + 1\right)\dfrac{\pi}{4}$

$S = \left\{\dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{7\pi}{4}\right\}$

For all four values of $\theta$, $\tan^{2}\theta \text{ and } co{\text{t}}^{2}\theta\text{will become}1\text{and}\sin^{2}\theta\text{will become}\dfrac{1}{2}$

So, ${x}^{2} - 2\left(\tan^{2}\theta + \cot^{2}\theta\right)x + 6\sin^{2}\theta = 0$

$⇒ {x}^{2} - 4x + 3 = 0$

Sum of roots of all four equations $= 4 \times 4 = 16$.