Animated Solution for Mathematics - Trigonometry: If $\theta \in [-2\pi ,2\pi ]$, then the number of solutions of $2\sqrt{2}{\cos }^2\theta +(2-\sqrt{6})\cos \theta -\sqrt{3}=0$, is equal to:

When you first encounter the equation

it is natural to feel a surge of anxiety. However, in the world of JEE Advanced, the most intimidating problems are often just simple concepts wearing a mask.

Let us peel back that mask. If you replace $\cos \theta$ with $x$, you are looking at the quadratic equation:

The secret to solving this efficiently lies in splitting the middle term. By expanding $(2-\sqrt{6})\cos \theta$, we obtain $2\cos \theta -\sqrt{6}\cos \theta$.

Substituting this back into the original equation, we get:

Now, we group the terms. From the first two terms, we factor out $2\cos \theta$, leaving us with $(\sqrt{2}\cos \theta +1)$. From the last two terms, we factor out $-\sqrt{3}$, which also leaves us with $(\sqrt{2}\cos \theta +1)$.

This yields the factored form:

This results in two fundamental cases:

We are working in the interval $\theta \in [-2\pi ,2\pi ]$. This represents two full cycles of the cosine wave.

For the case $\cos \theta =\frac{\sqrt{3}}{2}$, the horizontal line intersects the cosine wave twice in the positive cycle $[0,2\pi ]$ and twice in the negative cycle $[-2\pi ,0]$, yielding 4 solutions.

Similarly, for the case $\cos \theta =-\frac{1}{\sqrt{2}}$, the horizontal line again intersects the wave twice in each cycle, yielding another 4 solutions.

Summing these together, we arrive at a total of 8 solutions. With patience and structure, even the most intimidating problems crumble.