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

Given,

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

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

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

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

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

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

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

$\Rightarrow 2\theta =\left(2n+1\right)\frac{\pi }{2}$

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

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

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

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

$\Rightarrow x^2-4x+3=0$

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