Animated Solution for Mathematics - Inverse Trigonometric Functions: Let $S=\{x:{\cos }^{-1}x=\pi +{\sin }^{-1}x+{\sin }^{-1}(2x+1)\}$. Then ${\sum }_{x\in S}(2x-1{)}^2$ is equal to

Welcome, fellow math warriors! Today, we are going to dissect a problem that is a perfect example of why JEE Advanced is not just about knowing formulas—it is about having the discipline to respect the boundaries of the functions we work with.

We are looking at the equation:

At first glance, it looks like a mess of inverse trigonometric functions. But as we peel back the layers, you will see that it is actually a beautifully orchestrated dance of identities and algebraic reduction.

Our first instinct might be to panic at the sight of three different inverse trigonometric terms. But remember your toolkit! The most powerful weapon in your arsenal for this problem is the fundamental identity:

This identity is a bridge. It allows us to translate the language of cosine inverse into the language of sine inverse. By substituting ${\cos }^{-1}x=\frac{\pi }{2}-{\sin }^{-1}x$, our equation transforms into:

Now, we have a common language. Let us gather our terms. By moving the ${\sin }^{-1}x$ terms to the right, we isolate the most complex term:

Working with inverse functions directly is like trying to solve a puzzle with your eyes closed. Let us make it easier. Let ${\sin }^{-1}x=\theta$. This implies $x=\sin \theta$.

Now, our equation becomes:

To strip away the ${\sin }^{-1}$ on the left, we take the sine of both sides. Remember that $\sin (-\alpha )=-\sin \alpha$. Thus, we get:

Here, we use the allied angle formula: $\sin \left(\frac{\pi }{2}+A\right)=\cos A$. The equation simplifies elegantly to:

We are almost there. We need to return to the world of $x$. We know that $\cos 2\theta =1-2{\sin }^2\theta$. Since $x=\sin \theta$, we have $\cos 2\theta =1-2x^2$.

Substituting this back, we get:

Distributing the negative sign gives us $2x+1=-1+2x^2$. Rearranging everything to one side, we arrive at:

Dividing by 2, we find the quadratic equation: $x^2-x-1=0$. This is a classic quadratic, and its roots are given by the quadratic formula:

Now, stop. Do not rush to the final answer. In JEE Advanced, the most common way to lose marks is to ignore the domain.

The function ${\sin }^{-1}x$ is only defined for $x\in [-1,1]$. Similarly, ${\sin }^{-1}(2x+1)$ is only defined when $-1\le 2x+1\le 1$, which simplifies to $-1\le x\le 0$. The intersection of these two domains is $x\in [-1,0]$.

Let us look at our roots: $x_1=\frac{1+\sqrt{5}}{2}\approx 1.618$ and $x_2=\frac{1-\sqrt{5}}{2}\approx -0.618$. Clearly, $x_1$ is outside our valid interval of $[-1,0]$. We must reject it. The only valid solution is $x=\frac{1-\sqrt{5}}{2}$.

Finally, we calculate the requested sum: ${\sum }_{x\in S}(2x-1{)}^2$. Since $S$ contains only one element, we substitute $x=\frac{1-\sqrt{5}}{2}$:

And there it is! The complexity collapses into a simple integer. This problem taught us that while algebra provides the path, the domain constraints provide the map. Never ignore the map! The final answer is 5.