Problem 92 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The sum of all values of \(\theta \in\left(0, \frac{\pi}{2}\right)\) satisfying \(\sin ^{2} 2 \theta+\cos ^{4} 2 \theta=\frac{3}{4}\) is: \(\quad\) [Jan. 10,2019 (I)] (a) \(\pi\) (b) \(\frac{5 \pi}{4}\) (c) \(\frac{\pi}{2}\) (d) \(\frac{3 \pi}{8}\)

Step-by-Step Solution

Answer

There are no values of \(\theta\) in the given range satisfying the equation.

Step 1: Analysis of Given Equation

We start with the given equation: \(\sin^2 2\theta + \cos^4 2\theta = \frac{3}{4}\). Notice that \(\sin^2 x + \cos^2 x = 1\) implies \(\cos^4 x = (1 - \sin^2 x)^2\). Thus, the equation is about finding values of \(2\theta\).

Step 2: Substitution and Simplification

Let \(x = 2\theta\). Then the equation becomes \(\sin^2 x + \cos^4 x = \frac{3}{4}\). Substitute \(\cos^4 x = (1 - \sin^2 x)^2\) into the equation to get \(\sin^2 x + (1 - \sin^2 x)^2 = \frac{3}{4}\).

Step 3: Expand and Solve Quadratic

Expanding \(\cos^4 x\), we have \(1 - 2\sin^2 x + \sin^4 x\). The equation becomes \(\sin^2 x + 1 - 2\sin^2 x + \sin^4 x = \frac{3}{4}\), which simplifies to \(\sin^4 x - \sin^2 x + \frac{3}{4} = 0\).

Step 4: Use Substitution for Simplification

Let \(y = \sin^2 x\). The equation \(y^2 - y + \frac{3}{4} = 0\) is now a quadratic equation in \(y\).

Step 5: Solve Quadratic Equation

The quadratic equation is \(y^2 - y + \frac{3}{4} = 0\). Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a=1, b=-1, c=\frac{3}{4}\), gives the solutions for \(y\).

Step 6: Calculate Roots

Calculate the discriminant: \((-1)^2 - 4 \cdot 1 \cdot \frac{3}{4} = 1 - 3 = -2\), which is negative, showing that there are no real solutions. Thus no values of \(x\) satisfy the equation.

Step 7: Analyze for Mistakes in Process

Reassess the equation and solution path. Checking equations may reveal errors in assumption or setup. Either no solution exists in the required range, or hidden error in problem interpretation.

Key Concepts

trigonometric equations

Trigonometric equations involve finding angles that satisfy a specific trigonometric expression. These may include sine, cosine, tangent functions, among others. The exercise here involves finding values of \( \theta \) such that the equation \( \sin^2 2\theta + \cos^4 2\theta = \frac{3}{4} \) holds true within a given range. To solve these equations, a good starting point is to look at known trigonometric identities. Since \( \sin^2 x + \cos^2 x = 1 \), this can sometimes help simplify expressions or change the form of the equation. Substitution can also be a useful tool, such as setting \( x = 2\theta \) to reduce complexity and work with one variable instead of two. From there, careful algebraic manipulation helps isolate the trigonometric function and potentially solve for the angle.

quadratic equations in trigonometry

When a trigonometric equation simplifies to the form similar to a quadratic equation, it provides an interesting twist. This exercise showcases this when solving leads to \( \sin^4 x - \sin^2 x + \frac{3}{4} = 0 \). Here, a substitution \( y = \sin^2 x \) transforms it into \( y^2 - y + \frac{3}{4} = 0 \), which is a standard quadratic equation in \( y \). The quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is then used to find potential solutions for \( y \). However, solving this equation reveals no real roots, indicating no valid trigonometric solution for \( x \). Quadratic equations in trigonometry can thus show cases where no angles satisfy the initial trigonometric equation.

trigonometric identities

Trigonometric identities are essential tools in solving trigonometric equations. The identity \( \sin^2 x + \cos^2 x = 1 \) is fundamental and is utilized to transform or simplify equations. In the exercise, the identity is reorganized to express \( \cos^4 x \) as \( (1 - \sin^2 x)^2 \), helping break the equation into familiar parts. Using trigonometric identities strategically can reduce an equation to a solvable form, sometimes resulting in linear or quadratic equations. However, as shown in the solution process, solving might uncover cases where no real solutions exist, as indicated by a negative discriminant in the quadratic formula application. These identities are critical for expediting and clarifying the complex interplay of trigonometric functions eventually leading us to a logical conclusion.