Animated Solution for Mathematics - Trigonometry: If $0\le x<2\pi$, then the number of real values of $x$, which satisfy the equation $\cos x+\cos 2x+\cos 3x+\cos 4x=0$ is

Unlocking Trigonometric Equations

Welcome, future engineer. Today, we are not just solving an equation; we are learning to see the hidden architecture of mathematics.

When you first look at the equation $\cos x+\cos 2x+\cos 3x+\cos 4x=0$, it might feel like a chaotic jumble of waves. It is easy to feel overwhelmed by the prospect of expanding these terms into polynomials.

But pause. Take a breath. In the world of JEE Advanced, complexity is often just a mask for a beautiful, underlying symmetry.

The Art of Strategic Grouping

Imagine you are standing on a beach, watching four waves crash into each other. If you try to track every ripple individually, you will lose your mind. But if you look for the pattern, you see the rhythm.

Look at the arguments of our cosines: $x,2x,3x,$ and $4x$. Notice something? If we pair the first and the last, $x$ and $4x$, their sum is $5x$. If we pair the middle two, $2x$ and $3x$, their sum is also $5x$.

This is our 'Aha!' moment. We are not just adding numbers; we are aligning phases. By grouping them as $(\cos 4x+\cos x)+(\cos 3x+\cos 2x)=0$, we have transformed a four-term problem into a two-part harmony.

The Power of Identities

Now that we have our pairs, we need a tool to break them open. The sum-to-product identity is our scalpel here:

Let's apply this to our first group, $(\cos 4x+\cos x)$. Here, $A=4x$ and $B=x$. The sum is $5x$, and the difference is $3x$. Thus, we get $2\cos (\frac{5x}{2})\cos (\frac{3x}{2})$.

Now, for the second group, $(\cos 3x+\cos 2x)$. Here, $A=3x$ and $B=2x$. The sum is $5x$, and the difference is $x$. This yields $2\cos (\frac{5x}{2})\cos (\frac{x}{2})$.

Do you see it? The term $2\cos (\frac{5x}{2})$ has appeared in both expressions like a golden key. This is the reward for our strategic grouping.

The Factoring Dance

With our common factor identified, the equation becomes:

We are not done yet! We have a bracketed term that still contains a sum. Let's apply the sum-to-product identity one more time to $\cos (\frac{3x}{2})+\cos (\frac{x}{2})$.

Here, the sum of the angles is $\frac{3x}{2}+\frac{x}{2}=2x$, so the average is $x$. The difference is $\frac{3x}{2}-\frac{x}{2}=x$, so the average is $\frac{x}{2}$. This simplifies the bracket to $2\cos x\cos (\frac{x}{2})$.

Putting it all together, our equation has blossomed into:

The Final Harvest

Now, the equation is fully factored. For the product to be zero, at least one of the factors must be zero. This gives us three distinct cases to investigate within our interval $0\le x<2\pi$:

1. $\cos (\frac{5x}{2})=0$: This implies $\frac{5x}{2}=(2n+1)\frac{\pi }{2}$. Solving for $x$ in our range, we find $x=\frac{\pi }{5},\frac{3\pi }{5},\pi ,\frac{7\pi }{5},\frac{9\pi }{5}$.

2. $\cos x=0$: This gives $x=\frac{\pi }{2},\frac{3\pi }{2}$.

3. $\cos (\frac{x}{2})=0$: This gives $x=\pi$.

Wait! Look closely at our list. We found $x=\pi$ in the first case and again in the third case. We must be careful not to double-count.

When we collect our unique solutions: $\frac{\pi }{5},\frac{\pi }{2},\frac{3\pi }{5},\pi ,\frac{7\pi }{5},\frac{3\pi }{2},\frac{9\pi }{5}$, we count exactly seven distinct values.

We started with a daunting equation and ended with a clear, elegant set of solutions. This is the essence of JEE Advanced mathematics.

It is not about memorizing formulas; it is about recognizing patterns, applying the right tools with precision, and staying vigilant against traps like double-counting. You have successfully navigated this problem. Keep this mindset, and there is no equation you cannot conquer.