Animated Solution for Mathematics - Trigonometry: 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 $x^2-2({\tan }^2\theta +{\cot }^2\theta )x+6{\sin }^2\theta =0$, $heta\in S$, is ______.

Welcome, fellow traveler on the path to JEE excellence. Today, we are going to dismantle a trigonometric puzzle that, at first glance, looks like a chaotic mess of squared terms and double angles.

But remember, in the world of JEE Advanced, chaos is just order waiting to be discovered. Let us look at our given equation:

Our mission is to find the set $S$ of all $\theta$ in $(0,2\pi )$ that satisfy this. The secret here is unification. We have ${\cos }^2\theta$, ${\sin }^2\theta$, and ${\cos }^{2}2\theta$. We need to speak one language: the language of $\cos 2\theta$.

Watch closely as we perform a strategic split. We rewrite $7{\cos }^2\theta$ as $4{\cos }^2\theta +3{\cos }^2\theta$. Because now we have $3{\cos }^2\theta -3{\sin }^2\theta$, which is $3({\cos }^2\theta -{\sin }^2\theta )$, we recognize the identity:

Now our equation looks like this:

We are almost there, but that $4{\cos }^2\theta$ is still bothering us. We need it in terms of $\cos 2\theta$. Recall the identity $2{\cos }^2\theta =1+\cos 2\theta$. Therefore, $4{\cos }^2\theta =2(1+\cos 2\theta )$.

Substituting this back, we get:

Expanding the brackets, we have:

The constant $2$ on both sides cancels out beautifully, leaving us with:

This is a simple quadratic in $\cos 2\theta$. Factoring it, we get $\cos 2\theta (5-2\cos 2\theta )=0$. This gives us two possibilities: $\cos 2\theta =0$ or $\cos 2\theta =\frac{5}{2}$.

As we discussed, $\cos 2\theta =\frac{5}{2}$ is impossible, so we are left with $\cos 2\theta =0$.

If $\cos 2\theta =0$, then $2\theta$ must be an odd multiple of $\frac{\pi }{2}$. Since $\theta \in (0,2\pi )$, $2\theta$ ranges from $0$ to $4\pi$.

Thus, $2\theta =\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\frac{7\pi }{2}$. Dividing by $2$, we find our set $S$:

These are our four magical angles.

Now, we turn to the second part of our journey. We are given the equation:

For any $\theta \in S$, notice the symmetry. Whether $\theta$ is $\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},$ or $\frac{7\pi }{4}$, the value of ${\tan }^2\theta$ is always $1$, and ${\cot }^2\theta$ is also $1$. Similarly, ${\sin }^2\theta$ is always $\frac{1}{2}$.

This is the elegance of the problem! For every single angle in our set $S$, the quadratic equation becomes:

We have four identical quadratic equations, one for each $\theta$ in $S$. For the equation $x^2-4x+3=0$, the sum of the roots is:

Since we have four such equations, the total sum of all roots is $4\times 4=16$. We have navigated the trigonometry, conquered the quadratic, and arrived at the final answer of 16.