Problem 38 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Solution of the equation \(4 \sin ^{4} x+\cos ^{4} x=1\) is (A) \(x=n \pi\) (B) \(x=2 n \pi \pm \cos ^{-1}\left(\sqrt{\frac{3}{5}}\right)\) (C) \(x=(2 n+1) \frac{\pi}{2}\) (D) none of these

Step-by-Step Solution

Answer

A: The solution is \(x = n \pi\).

Step 1: Define and Simplify the given Equation

The equation provided is \(4 \sin^{4} x + \cos^{4} x = 1\). First, we will use the Pythagorean identity, \(\sin^2 x + \cos^2 x = 1\), to simplify the terms. Let \(a = \sin^2 x\) and \(b = \cos^2 x\), so we have \(a + b = 1\). Therefore, \(b = 1 - a\), and the equation becomes \(4a^2 + (1-a)^2 = 1\).

Step 2: Expand and Simplify the Quadratic Equation

Substitute \(b = 1-a\) into the equation: \(4a^2 + (1 - a)^2 = 1\) becomes \(4a^2 + 1 - 2a + a^2 = 1\). Simplifying gives \(5a^2 - 2a + 1 = 1\).

Step 3: Simplify Further and Solve for 'a'

Subtract 1 from both sides to get \(5a^2 - 2a = 0\). Factor out 'a': \(a(5a - 2) = 0\). This results in the solutions \(a = 0\) or \(5a - 2 = 0\).

Step 4: Solve for the Roots

From \(5a - 2 = 0\), solve to find \(a = \frac{2}{5}\). Now, we have \(\sin^2 x = 0\) or \(\sin^2 x = \frac{2}{5}\). Hence, \(\sin x = 0\) or \(\sin x = \pm \sqrt{\frac{2}{5}}\).

Step 5: Determine Solutions for x

For \(\sin x = 0\), the general solution is \(x = n\pi\). For \(\sin x = \pm \sqrt{\frac{2}{5}}\), the solution is \(x = n\pi + (-1)^n \sin^{-1}(\sqrt{\frac{2}{5}})\), which is equivalent to \(x = n\pi \pm \cos^{-1}(\sqrt{\frac{3}{5}})\) using the identity \(\sin^{-1} y = \frac{\pi}{2} - \cos^{-1} y\).

Step 6: Match Solutions to Options Given

Check the solutions obtained against the available options: Option (A), \(x = n\pi\), matches the condition when \(\sin x = 0\). None of the other options provided match the equivalent solutions.

Key Concepts

Pythagorean Identity

The Pythagorean Identity is a fundamental concept in trigonometry. It states that for any angle \(x\), \(\sin^2 x + \cos^2 x = 1\). This identity reflects the relationship between the sine and cosine functions on the unit circle.
Understanding this identity is crucial because it helps in simplifying trigonometric expressions. For instance, in the original exercise, we used the Pythagorean Identity to express \(\cos^2 x\) in terms of \(\sin^2 x\). We let \(a = \sin^2 x\) and \(b = \cos^2 x\), leading to the equation \(a + b = 1\). This means that \(b = 1 - a\).
The beauty of this identity lies in its simplicity and power. By transforming the original equation, \(4 \sin^4 x + \cos^4 x = 1\), using the Pythagorean Identity, we reduce the complexity of solving the problem.

• It allows for easy substitutions and simplifications.
• It’s applicable to any trigonometric equation involving sine and cosine.
• Helps in transforming equations into solvable forms.

With these transformations, solving complicated trigonometric equations becomes more manageable.

Quadratic Equation

Quadratic equations are polynomials of degree 2. In a typical form, they appear as \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. Solving a quadratic equation is a skill often required in mathematical problem-solving.
In the context of our trigonometric equation, once we used the Pythagorean identity, we introduced a simplified form, \(5a^2 - 2a + 1 = 1\). Subtracting 1 from both sides gave us the quadratic equation \(5a^2 - 2a = 0\). Solving quadratic equations often involves factoring, completing the square, or using the quadratic formula.

• We factored out \(a\) from the equation, resulting in \(a(5a - 2) = 0\).
• From the factored equation, we found potential solutions \(a = 0\) or \(5a - 2 = 0\).
• Solving \(5a - 2 = 0\) yields \(a = \frac{2}{5}\).

These solutions further led to solving \(\sin^2 x = 0\) and \(\sin^2 x = \frac{2}{5}\). These techniques highlight how quadratic equations are essential in breaking down and solving complex mathematical problems.

General Solution of Trigonometric Equations

The general solution of trigonometric equations refers to finding all possible solutions that satisfy the equation for any integer \(n\). For trigonometric functions, solutions often involve adding multiples of \(\pi\) or \(2\pi\), reflecting the periodic nature of the sine and cosine functions.
In our problem, once we found that \(\sin^2 x = 0\) or \(\sin^2 x = \frac{2}{5}\), we applied the concept of general solutions.

• For \(\sin x = 0\), the solutions are expressed as \(x = n\pi\), capturing the periodicity of the sine function where \(n\) is any integer.
• For \(\sin x = \pm \sqrt{\frac{2}{5}}\), we utilize the identity \(\sin^{-1} y = \frac{\pi}{2} - \cos^{-1} y\) to derive the solutions \(x = n\pi \pm \cos^{-1}(\sqrt{\frac{3}{5}})\).

These solutions ensure we account for all potential angles \(x\) that satisfy the original equation, considering the repetitive cycle of trigonometric functions. Understanding the general solution helps in covering all possible scenarios and ensuring no solutions are missed.