Animated Solution for Mathematics - Trigonometry: The number of solutions of the equation $\cos 2\theta \cos \frac{\theta }{2}+\cos \frac{5\theta }{2}=2{\cos }^3\frac{5\theta }{2}$ in $[-\frac{\pi }{2},\frac{\pi }{2}]$ is :

Welcome, fellow traveler of the mathematical landscape. Today, we stand before a trigonometric equation that, at first glance, seems like a tangled mess of angles:

It looks intimidating, doesn't it? But remember, in the world of JEE Advanced, intimidation is just a mask for elegance. Let's peel back the layers.

The Art of Grouping

When you see an equation like this, your instinct might be to expand everything. Resist that urge! The secret to trigonometry is often not expansion, but grouping.

Notice the term $\cos \frac{5\theta }{2}$ appearing on both sides. Let us bring them together. By moving the $\cos \frac{5\theta }{2}$ from the left to the right, we get:

Now, look at the right-hand side. It is begging to be factored. Pulling out $\cos \frac{5\theta }{2}$, we are left with:

The Identity Reveal

Does that bracket look familiar? It should! It is the classic double-angle identity in disguise: $2{\cos }^{2}A-1=\cos 2A$.

Here, our $A$ is $\frac{5\theta }{2}$. When we double that angle, we get $5\theta$. Suddenly, the equation transforms into something much cleaner:

The complexity has vanished, replaced by a beautiful symmetry.

The Product-to-Sum Transformation

We have products of cosines on both sides. To solve this, we need to break these products down using the product-to-sum formula: $2\cos A\cos B=\cos (A+B)+\cos (A-B)$.

To use it, we multiply the entire equation by $2$, giving us:

Applying the formula, the left side becomes:

The right side becomes:

The Final Simplification

Look at what we have created:

The term $\cos \frac{5\theta }{2}$ appears on both sides! We can cancel it out, leaving us with the elegant equation:

This is the moment of truth. We know that if $\cos X=\cos Y$, then $X=2n\pi \pm Y$. Applying this, we get:

The Boundary Hunt

We have two cases. Case 1 (the plus sign) gives us $\frac{15\theta }{2}=2n\pi +\frac{3\theta }{2}$, which simplifies to $6\theta =2n\pi$, or $\theta =\frac{n\pi }{3}$.

Case 2 (the minus sign) gives us $\frac{15\theta }{2}=2n\pi -\frac{3\theta }{2}$, which simplifies to $9\theta =2n\pi$, or $\theta =\frac{2n\pi }{9}$.

Now, we must respect our interval $[-\frac{\pi }{2},\frac{\pi }{2}]$. For $\theta =\frac{n\pi }{3}$, we find solutions at $0,\pm \frac{\pi }{3}$. For $\theta =\frac{2n\pi }{9}$, we find solutions at $0,\pm \frac{2\pi }{9},\pm \frac{4\pi }{9}$.

Counting these unique values, we find exactly 7 solutions. You have navigated the complexity, applied the identities, and respected the boundaries. That is the essence of JEE Advanced math—not just calculation, but the art of simplification.