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

Imagine standing before a complex trigonometric equation:

At first glance, it looks like a chaotic mess of different angles. We have $\theta$ and $2\theta$ fighting for dominance.

In the world of JEE Advanced, this is a classic setup designed to test your ability to find order in chaos. The secret to conquering this problem is not brute force, but finding a common language.

Our first mission is to unify the arguments. We cannot easily solve an equation that mixes $\theta$ and $2\theta$.

We need a bridge. That bridge is the double angle identity:

By substituting this into our equation, we transform the term $-10{\cos }^2\theta$ into $-5(2{\cos }^2\theta )$, which becomes $-5(1+\cos 2\theta )$. Suddenly, the equation breathes:

Now, watch the magic of algebra. When we expand the terms, we get:

The constants $-5$ and $+5$ vanish, leaving us with:

This is the moment of clarity. We have collapsed a terrifying expression into a simple, factorable quadratic form:

This gives us two distinct paths: $\cos 2\theta =0$ or $\cos 2\theta =-\frac{1}{3}$.

Here is where many students stumble. The original domain is $\theta \in [-4\pi ,4\pi ]$.

But our equation is in terms of $2\theta$. We must scale our domain. If $\theta$ is in $[-4\pi ,4\pi ]$, then $2\theta$ is in $[-8\pi ,8\pi ]$.

This is a total interval of $16\pi$, which spans exactly 8 periods of the cosine function.

For $\cos 2\theta =0$, the graph crosses the axis twice in every $2\pi$ period. With 8 periods, that is $8\times 2=16$ solutions.

For $\cos 2\theta =-\frac{1}{3}$, the horizontal line $y=-\frac{1}{3}$ intersects the curve twice in every $2\pi$ period. Again, $8\times 2=16$ solutions.

Adding these together, we arrive at a total of 32 solutions. It is a beautiful, symmetrical result.

Remember, in mathematics, the most complex problems often have the most elegant solutions if you are willing to look for the underlying structure.