Animated Solution for Mathematics - Trigonometry: The number of solutions of ${\sin }^{7}x+{\cos }^{7}x=1,x\in [0,4\pi ]$ is equal to

Facing the Power of Seven

The problem, ${\sin }^{7}x+{\cos }^{7}x=1$, looks like a jagged, impossible peak. That power of $7$ is a psychological trap, designed to make you reach for complex expansions or tedious substitutions.

But in the world of JEE Advanced, the most intimidating problems often have the most elegant, simple hearts. We are not here to fight the power of $7$; we are here to outsmart it.

Our stage is the interval $x\in [0,4\pi ]$. We are looking for the points where the graph of this function kisses the horizontal line $y=1$.

The Bounding Technique

To conquer this, we need a benchmark. We know the golden rule of trigonometry:

This identity is our anchor. Now, consider the behavior of $\sin x$. We know that for any real $x$, $-1\le \sin x\le 1$.

When you take a value in this range and raise it to a power, something interesting happens. If you take a fraction like $0.5$ and square it, you get $0.25$. If you raise it to the power of $7$, you get $0.0078125$.

The value shrinks! Mathematically, for any $a$ such that $|a|\le 1$, we have $a^n\le a^2$ for any $n\ge 2$. This is the secret weapon. It means that ${\sin }^{7}x\le {\sin }^{2}x$ and ${\cos }^{7}x\le {\cos }^{2}x$.

Adding the Pieces

Now, let's bring these two inequalities together. If we add them, we get:

Since the right side is exactly $1$, we have proven that ${\sin }^{7}x+{\cos }^{7}x\le 1$. This is a massive breakthrough. It tells us that the function can never rise above the line $y=1$.

The only way the equation ${\sin }^{7}x+{\cos }^{7}x=1$ can be satisfied is if the inequality becomes an equality. This happens only when ${\sin }^{7}x={\sin }^{2}x$ and ${\cos }^{7}x={\cos }^{2}x$ simultaneously.

Solving the Conditions

Let's solve ${\sin }^{7}x={\sin }^{2}x$. This rearranges to:

This gives us two possibilities: $\sin x=0$ or $\sin x=1$. Similarly, for ${\cos }^{7}x={\cos }^{2}x$, we get $\cos x=0$ or $\cos x=1$.

But wait, we must respect the identity ${\sin }^{2}x+{\cos }^{2}x=1$. They cannot both be $1$, and they cannot both be $0$. This leaves us with two clean, distinct cases:

Case 1: $\sin x=0$ and $\cos x=1$. Case 2: $\sin x=1$ and $\cos x=0$.

Mapping the Solutions

Let's look at Case 1: $\sin x=0$ and $\cos x=1$. This happens at the start of the unit circle, at $x=0$, and repeats every $2\pi$. In our interval $[0,4\pi ]$, the solutions are $x=0,2\pi ,4\pi$.

Now, Case 2: $\sin x=1$ and $\cos x=0$. This happens at the top of the unit circle, at $x=\frac{\pi }{2}$, and repeats every $2\pi$. In our interval, the solutions are $x=\frac{\pi }{2},\frac{5\pi }{2}$.

Counting them all up, we have $0,\frac{\pi }{2},2\pi ,\frac{5\pi }{2},4\pi$. That is exactly 5 solutions.

You see? The power of $7$ was just a mask. By using the bounding technique, we stripped away the complexity and found the truth hidden underneath. Keep this technique in your toolkit—whenever you see high powers in trigonometry, think about how they are bounded by their squares.