Animated Solution for Mathematics - Trigonometry: The equation $2{\cos }^2\frac{x}{2}{\sin }^{2}x=x^2+x^{-2}$; $0<x\le \frac{\pi }{2}$ has

Imagine you are standing on the edge of a mathematical cliff. You are looking at the equation:

The domain is defined as $0<x\le \frac{\pi }{2}$. Your first instinct might be to reach for your algebraic toolkit—to expand, to rearrange, or to solve for $x$.

However, in the world of JEE Advanced, sometimes the most powerful move is not to solve, but to observe.

Many students fall into the trap of trying to force a solution. They see $x^2+\frac{1}{x^2}$ and think of differentiation or squaring, while the trigonometric side suggests complex identities.

But this is a transcendental equation. It represents a collision between two different worlds: the periodic, bounded world of trigonometry and the unbounded, growing world of algebra.

When you see this, do not solve. Instead, ask: "What is the maximum possible value of the left side, and what is the minimum possible value of the right side?" This is the bounding strategy.

Let us look at the Left Hand Side (LHS): $2{\cos }^2\left(\frac{x}{2}\right){\sin }^{2}x$. We can use the half-angle identity $2{\cos }^2\left(\frac{\theta }{2}\right)=1+\cos \theta$.

Substituting $\theta =x$, the LHS transforms into:

Now, consider the interval $0<x\le \frac{\pi }{2}$. In this domain, $\cos x$ is always strictly less than $1$, meaning $(1+\cos x)<2$.

Furthermore, we know that ${\sin }^{2}x$ can never exceed $1$. When you multiply a number strictly less than $2$ by a number at most $1$, the result must be strictly less than $2$.

The LHS is trapped below the value of $2$.

Now, turn your gaze to the Right Hand Side (RHS): $x^2+x^{-2}$, which is $x^2+\frac{1}{x^2}$. Here, the Arithmetic Mean-Geometric Mean (AM-GM) inequality is your best friend.

For any positive real number $a$, $a+\frac{1}{a}\ge 2$. Applying this to $x^2$, we get:

The minimum value of the RHS is $2$, occurring exactly when $x^2=\frac{1}{x^2}$, or $x=1$. The RHS is anchored at $2$ and can only grow larger as $x$ moves away from $1$.

We have reached the climax of our journey. The LHS is always strictly less than $2$, while the RHS is always greater than or equal to $2$.

Like two ships passing in the night, they can never touch. There is no value of $x$ in the given domain where the LHS can reach the RHS.

Therefore, the equation has no real solution. This result is a triumph of logical deduction; you did not need to solve for $x$, but simply understand the nature of the functions involved.