Problem 54 - A Complete Resource Book in Mathematics for JEE Main

Question

Solution of the equation \(\sin ^{-1} x+\sin ^{-1} 2 x=\frac{\pi}{3}\) is (A) \(x=\frac{\sqrt{3}}{2 \sqrt{7}}\) (B) \(x=-\frac{\sqrt{3}}{2 \sqrt{7}}\) (C) \(x=\pm \frac{1}{\sqrt{2}}\) (D) none of these

Step-by-Step Solution

Answer

The solution is (A) \(x=\frac{\sqrt{3}}{2\sqrt{7}}\).

Step 1: Set up the Equation

We start with the given equation: \( \sin^{-1} x + \sin^{-1} 2x = \frac{\pi}{3} \). The goal is to find the value of \( x \).

Step 2: Apply the Identity

Recall the identity: \( \sin^{-1} a + \sin^{-1} b = \sin^{-1} (a\sqrt{1-b^2} + b\sqrt{1-a^2}) \) for \( a^2 + b^2 \leq 1 \). Here, \( a = x \) and \( b = 2x \).

Step 3: Substitute into the Identity

Substitute \( a = x \) and \( b = 2x \) into the identity: \( \sin^{-1} x + \sin^{-1} 2x = \sin^{-1} (x\sqrt{1-(2x)^2} + 2x\sqrt{1-x^2}) \).

Step 4: Simplify Inside the Sine Inverse

Calculate the expression inside the sine inverse: \( x\sqrt{1-4x^2} + 2x\sqrt{1-x^2} \). Simplify this to get \( x(\sqrt{1-4x^2} + 2\sqrt{1-x^2}) \).

Step 5: Equate and Solve

Set \( \sin^{-1} (x(\sqrt{1-4x^2} + 2\sqrt{1-x^2})) = \frac{\pi}{3} \). Solve for \( x \) using trigonometric identities, such as converting the sine inverse equation to sine, \( \sin(\sin^{-1}(x(..))) = \sin(\frac{\pi}{3}) \).

Step 6: Solve for x

The equation becomes a quadratic: \( x(\sqrt{1-4x^2} + 2\sqrt{1-x^2}) = \frac{\sqrt{3}}{2} \). Solve this equation by squaring and simplifying terms to find potential solutions for \( x \).

Step 7: Check Possible Solutions

From simplification, derive potential solutions like \( x = \frac{\sqrt{3}}{2\sqrt{7}} \) and \( x = -\frac{\sqrt{3}}{2\sqrt{7}} \). Validity of solutions depends on conditions in the domain where \( \sin^{-1}(..) + \sin^{-1}(..) \) is defined.

Key Concepts

Sine Inverse Function

The sine inverse function is crucial when dealing with inverse trigonometric equations. It is denoted as \( \sin^{-1}(x) \), and it represents the angle whose sine is \( x \). The range for sine inverse values is typically between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\). This constraint is important as it ensures a unique solution for each input.When solving equations with multiple sine inverse terms, conversion to a direct trigonometric function can simplify the process.

• Remember: \( \sin(\sin^{-1}(x)) = x \).
• It is important that the value of \( x \) in the expression \( \sin^{-1}(x) \) lies between -1 and 1, because these are the bounds for the sine function.

Breaking down expressions involving \( \sin^{-1} \) can often reveal simpler forms or identities that help in solving the equation.

Trigonometric Identities

Trigonometric identities are equations that are true for all values of the variables involved. They're often used to simplify or transform expressions. In this exercise, a relevant identity is:\[\sin^{-1}(a) + \sin^{-1}(b) = \sin^{-1}\left( a\sqrt{1-b^2} + b\sqrt{1-a^2} \right)\]This identity is especially useful when solving equations that involve sums of inverse sine terms. When using this identity:

• Ensure \( a^2 + b^2 \leq 1 \) for the identity to hold true.
• The equation helps transform complex trigonometric expressions into ones that are more straightforward to handle.

By substituting the variables from the exercise into this identity, we obtained a new expression that could be further simplified and solved.

Quadratic Equations

Quadratic equations are polynomials of the form \( ax^2 + bx + c = 0 \). Solving them often involves factoring, completing the square, or using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In the context of this exercise, transitioning from a trigonometric identity to a quadratic equation was key to finding the solution. Simplifying the terms inside the sine inverse using algebra and identities helped:

• Identify coefficients and constants to form the quadratic equation.
• Use algebraic manipulation like squaring both sides to facilitate solving.

Checking solutions against the domain of the original inverse function guarantees the validity of your answers. Quadratic equations resulting from trigonometric problems may also require consideration of trigonometric constraints to determine the acceptable solution set.