Animated Solution for Mathematics - Trigonometry: The real roots of the equation ${\cos }^{7}x+{\sin }^{4}x=1$ in the interval $(-\pi ,\pi )$ are ........., ........., and ..........

Imagine you are standing on the unit circle, looking at the equation ${\cos }^{7}x+{\sin }^{4}x=1$. It looks innocent enough, but it hides a beautiful, rigid structure.

We are tasked with finding all real roots in the open interval $(-\pi ,\pi )$. When you see high powers of trigonometric functions, your first instinct might be to panic, but let's pause and look at the constraints.

Both $\cos x$ and $\sin x$ are trapped between $-1$ and $1$. This is our key.

Let's start by rearranging the equation to group the terms. If we subtract ${\sin }^{4}x$ from both sides, we get ${\cos }^{7}x=1-{\sin }^{4}x$.

Now, look at the right-hand side. It screams for the difference of squares identity: $a^2-b^2=(a-b)(a+b)$. We can rewrite $1-{\sin }^{4}x$ as $(1-{\sin }^{2}x)(1+{\sin }^{2}x)$.

Using the fundamental identity ${\sin }^{2}x+{\cos }^{2}x=1$, we know that $1-{\sin }^{2}x={\cos }^{2}x$. Substituting this back, our equation becomes:

Here is where many students stumble. They see ${\cos }^{2}x$ on both sides and are tempted to divide. But wait! If $\cos x=0$, you would be dividing by zero, which is a mathematical sin.

Instead, let's bring everything to one side:

Now, we can safely factor out ${\cos }^{2}x$:

This gives us two distinct paths to explore.

This is straightforward. If ${\cos }^{2}x=0$, then $\cos x=0$. Within our interval $(-\pi ,\pi )$, the cosine function hits zero at $x=\frac{\pi }{2}$ and $x=-\frac{\pi }{2}$. These are our first two roots.

This looks intimidating, but let's use bound analysis. We know that $\cos x$ is at most $1$, so ${\cos }^{5}x\le 1$.

On the other side, ${\sin }^{2}x$ is always non-negative, so $1+{\sin }^{2}x\ge 1$. The only way for a value that is at most $1$ to equal a value that is at least $1$ is if both sides are exactly $1$.

This forces ${\cos }^{5}x=1$ (which means $\cos x=1$) and $1+{\sin }^{2}x=1$ (which means $\sin x=0$). The only angle in our interval that satisfies both $\cos x=1$ and $\sin x=0$ is $x=0$.

We have found our three roots: $-\frac{\pi }{2},0,\text{ and }\frac{\pi }{2}$. Each one fits perfectly within our interval.

By resisting the urge to divide and instead using the power of factorization and bound analysis, we have tamed this trigonometric beast. Remember, in JEE Advanced, it is rarely about brute force; it is about finding the elegant path through the forest of equations.