Animated Solution for Mathematics - Trigonometry: The number of solutions of the equation $\cos \left(x+\frac{\pi }{3}\right)\cos \left(\frac{\pi }{3}-x\right)=\frac{1}{4}{\cos }^{2}2x$, $x\in [-3\pi ,3\pi ]$ is :

Welcome, future engineer. Today, we are going to dismantle a problem that looks intimidating but is actually a masterclass in elegance.

When you first see the equation

your brain might immediately want to expand the terms using the addition formulas. Stop. Take a breath. In JEE Advanced, the most complex-looking problems often have the simplest solutions if you can spot the hidden pattern.

The Identity

Look closely at the left-hand side: $\cos (x+\frac{\pi }{3})\cos (\frac{\pi }{3}-x)$. This is a classic structure. It is the product of two cosines where the arguments are symmetric.

We have a powerful tool for this: the identity

By setting $A=\frac{\pi }{3}$ and $B=x$, the left side collapses beautifully into ${\cos }^2(\frac{\pi }{3})-{\sin }^{2}x$.

Since $\cos (\frac{\pi }{3})=\frac{1}{2}$, we know that ${\cos }^2(\frac{\pi }{3})=\frac{1}{4}$. Our equation is now:

See how the chaos is already fading?

The Unification

Now, we have a mix of $x$ and $2x$. We cannot solve this until we speak the same language. We need to convert everything to $2x$.

We know the double-angle identity ${\sin }^{2}x=\frac{1-\cos 2x}{2}$. Let's multiply the entire equation by $4$ first to clear those fractions:

Now, substitute our identity:

Expanding this gives us $1-2+2\cos 2x={\cos }^{2}2x$, which simplifies to:

The Quadratic Revelation

Bring everything to one side, and you will see the magic happen:

This is not just any equation; it is a perfect square! It is:

This implies that $\cos 2x=1$. This is the moment where the problem yields. We are no longer dealing with complex identities; we are simply solving for when the cosine function equals one.

The Final Count

The general solution for $\cos \theta =1$ is $\theta =2n\pi$. Therefore, $2x=2n\pi$, which simplifies to $x=n\pi$, where $n$ is any integer.

We are looking for solutions in the interval $x\in [-3\pi ,3\pi ]$. Substituting our general solution, we get:

Dividing by $\pi$, we find that $n$ must be an integer such that $-3\le n\le 3$. The possible values for $n$ are $\{-3,-2,-1,0,1,2,3\}$.

Counting these, we find exactly 7 solutions. You have successfully navigated the trap and arrived at the truth. Keep this mindset—look for the identity, unify the variables, and simplify. You are ready for anything.