Problem 23 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Solve the following trigonometric equations: If the equation \(\sin ^{4} x-(k+2) \sin ^{2} x-(k+3)=0\) has a solution, then find the value of \(k\).

Step-by-Step Solution

Answer

The solution exists when \(k \leq -2\sqrt{2}\) or \(k \geq 2\sqrt{2}\).

Step 1: Identify the Quadratic Form

Recognise that the given trigonometric equation \(\sin ^{4} x-(k+2) \sin ^{2} x-(k+3)=0\) can be seen as a quadratic equation \(ax^{2} + bx + c = 0\), where x is replaced by \(\sin^{2}x\). Therefore, we can rewrite the equation as \(u^{2} -(k+2)u -(k+3) = 0\), where \(u = \sin^{2}x\).

Step 2: Identify a, b and c

In our quadratic equation \(u^{2} - (k+2)u - (k+3) = 0\), we can identify our a, b and c as follows: a = 1 (coefficient of \(u^{2}\)), b = -(k+2) (coefficient of u), and c = -(k+3) (the constant term).

Step 3: Evaluate the Discriminant

The discriminant, typically denoted D, is used to identify the nature of roots of a quadratic equation and is given by \(b^{2} - 4ac\). Substituting b = -(k+2), a = 1 and c = -(k+3), we get \(D = [-(k+2)]^{2} - 4*1*[-(k+3)] = (k+2)^{2} - 4(k+3)\).

Step 4: Determine the Condition for D

We know that for a quadratic equation to possess real roots (i.e., to have a solution), the discriminant should be greater than or equal to 0. Thus, the inequality we need to solve is \((k+2)^{2} - 4(k+3) \geq 0 \).

Step 5: Solve for k

Solving \((k+2)^{2} - 4(k+3) \geq 0\) would result in the range of values of k that satify the condition. Expanding and simplifying gives \(k^{2}+4k+4-4k-12 \geq 0 \Rightarrow k^{2}-8 \geq 0.\) Solving this inequality gives \(k \leq -2\sqrt{2}\) or \(k \geq 2\sqrt{2}.\)

Key Concepts

Quadratic Equations

Quadratic equations are a cornerstone of algebra and are widely encountered in various math problems. A quadratic equation typically takes the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable to be solved for. Quadratic equations are known for having either two, one, or zero real solutions depending on the context.

These solutions can be found using methods such as:

• Factoring
• Using the quadratic formula
• Completing the square

Interestingly, trigonometric equations like the one presented in the problem can sometimes be rewritten into a quadratic form, allowing us to apply these methods. In this exercise, the substitution \(u = \sin^{2}x\) turned the trigonometric equation into the quadratic form \(u^2 -(k+2)u -(k+3) = 0\). This helps simplify the problem to something more manageable.

Once in quadratic form, we can identify the coefficients \(a\), \(b\), and \(c\) to further explore the equation.

Discriminant in Quadratics

The discriminant is a vital element in analyzing the roots of a quadratic equation. Given the standard form \(ax^2 + bx + c = 0\), the discriminant \(D\) is defined as \(b^2 - 4ac\). This value determines the nature of the roots without having to solve the entire equation.

Here's what the discriminant reveals about the equation's solutions:

• \(D > 0\): Two distinct real roots exist.
• \(D = 0\): There is exactly one real root (a repeated root).
• \(D < 0\): No real roots; solutions are complex numbers.

In our problem, the equation being in the form \(u^2 -(k+2)u -(k+3) = 0\) means we calculate the discriminant as \((k+2)^2 - 4(-k-3)\). For the quadratic to have real solutions, \(D\) must be \(\geq 0\). Hence, the inequality \((k+2)^2 - 4(k+3) \geq 0\) is solved to determine suitable values for \(k\) that ensure the presence of real solutions.

Trigonometric Functions

Trigonometric functions are fundamental in connecting angles to side lengths in right triangles. The core trigonometric functions include sine, cosine, and tangent, each of which corresponds to a ratio of sides in a triangle.

In solving trigonometric equations, we often seek values that satisfy the equation within a specific interval, typically using known values of these functions. For instance, the equation in our exercise involves \(\sin(x)\), specifically examining \(\sin^{2}(x)\), a common transformation that simplifies trigonometric solutions.

Transformations like \( \sin^{4}(x) \) into a quadratic expression using \( u = \sin^{2}(x) \) are powerful tools. They allow us to treat such trigonometric equations much like standard algebraic equations applicable for quadratic solutions. This method simplifies the complexity of direct trigonometric solutions, granting access to elegant algebraic techniques that can be applied universally across various trigonometric problems.