Problem 17 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Solve for \(x: \sin ^{-1} x+\sin ^{-1} 2 x=\frac{\pi}{2}\).

Step-by-Step Solution

Answer

The solution is \(x= -\frac{\sqrt{10}}{5}\).

Step 1: Apply Indirect trigonometric function

If we apply the indirect trigonometric function to both sides, we get that: \( \sin ( \sin^{-1} x + \sin^{-1} 2x) = \sin \left( \frac{\pi}{2} \right) \). This simplifies to \( \sin ( \sin^{-1} x + \sin^{-1} 2x) = 1 \) as the sine of \( \frac{\pi}{2} \) is 1.

Step 2: Use the formula of sine of sum of two terms

One can use the formula for sine of sum of two terms to the left side of the equation. This formula is \( \sin (a+b) = \sin a \cos b + \cos a \sin b \) . This leaves us with \(x \sqrt{1-4x^2} + \sqrt{1-x^2} 2x = 1\).

Step 3: Square both sides

At this point one can square both sides of the equation to remove the root elements. This gives us \((x \sqrt{1-4x^2} + \sqrt{1-x^2} 2x)^2 = 1^2\) which simplifies to \(x^2(1-4x^2) + 4x^2(1-x^2) + 4x^2 \sqrt{(1-4x^2)(1-x^2)} = 1\).

Step 4: Ignore the root term

The last term under the square root is difficult to deal with, so one can ignore it and solve for this equation first: \(x^2(1-4x^2) + 4x^2(1-x^2) = 1\). After simplifying we get \(5x^4 - 2x^2 + 1 = 0\).

Step 5: Substitute variables and solve quadratic equation

We substitute \(y = x^2\). Therefore, we have \(5y^2 - 2y + 1 = 0\). Solving this quadratic equation, we have the roots \(y_1 = y_2 = \frac{2}{5}\). So, \(x^2 = \frac{2}{5}\). Solving this we arrive at \(x =\pm \frac{\sqrt{10}}{5}\).

Step 6: Check the solutions

Insert these two solutions back into the original equation to verify. Only \(-\frac{\sqrt{10}}{5}\) is suitable in the original equation.