Animated Solution for Mathematics - Inverse Trigonometric Functions: Let $S$ be the set of all solutions of the equation ${\cos }^{-1}(2x)-2{\cos }^{-1}(\sqrt{1-x^2})=\pi ,x\in [-\frac{1}{2},\frac{1}{2}]$. Then ${\sum }_{x\in S}2{\sin }^{-1}(x^2-1)$ is equal to

Welcome, my dear student. Today, we are not just solving a problem; we are embarking on a journey through the elegant landscape of inverse trigonometry.

When you first look at the equation ${\cos }^{-1}(2x)-2{\cos }^{-1}(\sqrt{1-x^2})=\pi$, it might seem intimidating. It feels like a tangled knot of functions.

But remember, every complex problem in JEE Advanced is just a series of simple, logical steps waiting to be uncovered. Let us peel back the layers together.

The Domain Gatekeeper

Before we touch a single variable, we must respect the boundaries. The problem explicitly states $x\in [-\frac{1}{2},\frac{1}{2}]$.

This is not just a suggestion; it is the law of this land. Many students rush into the algebra, solve the quadratic, and then get trapped by extraneous roots.

By keeping this interval in our minds, we are already ahead of the curve.

The Identity Toolkit

Now, look at the second term: $2{\cos }^{-1}(\sqrt{1-x^2})$. It screams for simplification.

We have a powerful tool in our arsenal: the identity $2{\cos }^{-1}(u)={\cos }^{-1}(2u^2-1)$, valid for $u\in [0,1]$. By setting $u=\sqrt{1-x^2}$, we can transform this complex term into something much more manageable.

Let us perform the substitution:

Squaring the square root is a delight—it gives us $1-x^2$. Multiplying by $2$ and subtracting $1$, we arrive at $2(1-x^2)-1=1-2x^2$.

Just like that, the term has simplified to ${\cos }^{-1}(1-2x^2)$.

The Algebraic Leap

With our simplified term, the original equation transforms into:

This is the turning point. We want to eliminate the inverse trigonometric functions. Let us rearrange the terms to isolate the functions:

Now, we apply the cosine function to both sides. On the left, we use the property $\cos (\theta -\pi )=-\cos (\theta )$.

Let $\theta ={\cos }^{-1}(2x)$. The left side becomes $-\cos ({\cos }^{-1}(2x))$, which is simply $-2x$.

On the right side, the cosine and inverse cosine functions neutralize each other, leaving us with $1-2x^2$. We have successfully transitioned from the abstract world of trigonometry to the concrete world of algebra:

The Quadratic Reality

We are left with a standard quadratic equation: $2x^2-2x-1=0$. Using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, we find the roots:

Now, we return to our gatekeeper. We have two candidates: $x_1=\frac{1+\sqrt{3}}{2}\approx 1.366$ and $x_2=\frac{1-\sqrt{3}}{2}\approx -0.366$.

The first root is clearly outside our interval $[-\frac{1}{2},\frac{1}{2}]$, so we reject it. The second root, $x_2$, fits perfectly.

We are almost there. The question asks for $2{\sin }^{-1}(x^2-1)$. With only one valid $x$, we calculate $x^2$:

Thus, $x^2-1=-\frac{\sqrt{3}}{2}$. Finally, we evaluate $2{\sin }^{-1}(-\frac{\sqrt{3}}{2})$.

Since ${\sin }^{-1}(-\frac{\sqrt{3}}{2})=-\frac{\pi }{3}$, our final answer is:

You see? By staying calm and following the logic, we turned a terrifying equation into a beautiful, simple result. Keep this confidence, and you will conquer any problem that comes your way.