Problem 13 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Find the number of solution of the equation \(2 \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=\pi x^{3}\)

Step-by-Step Solution

Answer

The equation has 3 solutions: \(-\sqrt{2},0,\sqrt{2}\)

Step 1: Transform the given equation

We have the equation \(2 \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=\pi x^{3}\). To simplify, let's set \(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) = a\). Therefore, our equation is reduced to the linear equation \(2a = \pi x^{3}\).

Step 2: Remember the trigonometric identity

The identity \(\sin (2 a) = 2 \sin a \cos a\) allows us to find \(\cos a\). Now, \(\sin a = \frac{2 x}{1+x^{2}}\) and \(\cos a = \sqrt{1-\sin^2 a} = \sqrt{1-\left(\frac{2 x}{1+x^{2}}\right)^2} = \frac{1-x^{2}}{1+x^{2}}\).

Step 3: Solve for x

We know that the function \(\sin a\) can only have values in the interval [-1,1]. So we need to find the values of \(x\) for which \(\frac{2 x}{1+x^{2}}\) is within the interval [-1,1]. This gives us the intervals of \(x\) as \(-1\sqrt{2}\le x \le -\sqrt{2}\) or \(\sqrt{2}\le x \le \sqrt{2}\). Using these intervals, we find that the solutions for \(x\) are \(-\sqrt{2},0,\sqrt{2}\).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions help us find angles when we know the trigonometric ratio. In this exercise, we see \( \sin^{-1} \left( \frac{2x}{1+x^2} \right) \), which means we are looking for an angle whose sine value is \( \frac{2x}{1+x^2} \). This is useful for solving equations when the trigonometric ratio is known, but the angle is unknown.

Inverse sine, for example, maps ratios back to angles typically between \[ -\frac{\pi}{2}, \frac{\pi}{2} \]. Reflect that inverse sine gives responses within this interval because sine only reaches all values from -1 to 1 here.

Always be cautious with inverse trigonometric functions, as their output might be limited to certain intervals, affecting solutions you find.

Trigonometric Identities

Trigonometric identities are formulas involving trigonometric functions that hold true for all values of the angle where they are defined.

In this exercise, we used the identity \( \sin(2a) = 2 \sin(a) \cos(a) \) to work with the transformed equation. This identity helped us here to express \( \sin(2a) \) in terms of \( \sin(a) \) and \( \cos(a) \), making calculations possible.

These identities allow us to manipulate trigonometric equations and find solutions more easily. They come in handy when simplifying expressions, proving trigonometric equations, or integrating functions.

Interval Analysis

Interval analysis involves finding values within specified ranges where a function behaves in a certain way. In this exercise, we focus on finding valid \( x \) that fit within required constraints.

We ensure that the expression \( \frac{2x}{1+x^2} \) fits within \([-1, 1] \) because it must be a valid input for the inverse sine function. Solving where this holds true reveals the solution intervals:

• \(-\sqrt{2} \le x \le 0\)
• \( \sqrt{2} \ge x \ge 0 \)

By determining these intervals, we identify the points where the original equation is solvable, thus pinpointing possible solutions. Ensuring correct intervals guarantees we don't pick impossible or extraneous solutions.