The number of elements in the set $S=\theta \in \left[-4\pi ,4\pi \right]: 3{\cos }^{2}2\theta +6\cos 2\theta -10{\cos }^2\theta +5=0$ is ______.

$3{\cos }^{2}2\theta +6\cos 2\theta -10{\cos }^2\theta +5=0$

$3{\cos }^{2}2\theta +6\cos 2\theta -5\left(1+\cos 2\theta \right)+5=0$

$\cos 2\theta \left(3\cos 2\theta +1\right)=0$

$\cos 2\theta =0$ or $\cos 2\theta =-\frac{1}{3}$

Given $\theta \in \left[-4\pi ,4\pi \right]\Rightarrow 2\theta \in \left[-8\pi ,8\pi \right]$

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

$\theta =\pm \frac{\pi }{4},\pm \frac{3\pi }{4},\dots \dots \dots \pm 15\frac{\pi }{4}$ gives $16$ solution

Similarly $\cos 2\theta =-\frac{1}{3}$ gives $16$ solution

hence, total elements in set $S$ will be $32$