Problem 33 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(|\cos x|^{\sin ^{2} x-\frac{3}{2} \sin x+\frac{1}{2}}=1\), then possible values of \(x\) are (A) \(n \pi\) or \(n \pi+(-1)^{n} \frac{\pi}{6}, n \in I\) (B) \(n \pi\) or \(2 n \pi+\frac{\pi}{2}\) or \(n \pi+(-1)^{n} \frac{\pi}{6}, n \in I\) (C) \(n \pi+(-1)^{n} \frac{\pi}{6}, n \in I\) (D) \(n \pi, n \in I\)

Step-by-Step Solution

Answer

The possible values of \(x\) are given by option (A).

Step 1: Understand the Equation

We start with the equation \(|\cos x|^{\sin ^{2} x-\frac{3}{2} \sin x+\frac{1}{2}}=1\). This means the base \(|\cos x|\) raised to the power of \(\sin^2 x - \frac{3}{2} \sin x + \frac{1}{2}\) equals 1.

Step 2: Analyze When a Power Equals One

An expression in the form \(a^b = 1\) can be true if either \(a = 1\), \(a = -1\) with an even \(b\), or \(b = 0\). We will explore these cases.

Step 3: Case a = 1, a = |cos x|

For \(|\cos x| = 1\), \(\cos x\) must either be 1 or -1. The cosine function equals 1 at \(x = 2n\pi\) and -1 at \(x = (2n+1)\pi\). Therefore, \(x = n\pi, n\in I\).

Step 4: Verify b = 0

Set \(b = \sin^2 x - \frac{3}{2} \sin x + \frac{1}{2} = 0\). Solve the quadratic equation by setting \( y = \sin x\): \[ y^2 - \frac{3}{2}y + \frac{1}{2} = 0 \] Using the quadratic formula, roots are \( y = 1 \) or \( y = \frac{1}{2} \). Thus, \(\sin x = 1\) or \(\sin x = \frac{1}{2}\).

Step 5: Solve for x in sin x = 1

\(\sin x = 1\) occurs at \(x = (2n+1)\frac{\pi}{2}, n \in I\). However, these values are not covered explicitly in the multiple-choice options, suggesting reviewing other solutions.

Step 6: Solve for x in sin x = 1/2

\(\sin x = \frac{1}{2}\) at \(x = n\pi + (-1)^n\frac{\pi}{6}\), which corresponds to the angles \(\frac{\pi}{6}\) or \(\frac{5\pi}{6}\) when adding multiples of \(\pi\).

Step 7: Combine Possible Solutions

Combining our solutions: For \(|\cos x| = 1\), \(x = n\pi\). For \(b = 0\), \(x = n\pi + (-1)^n \frac{\pi}{6}\). These solutions match option (A) completely.

Key Concepts

Trigonometric Equations

Trigonometric equations like \(|\cos x|^{\sin^{2} x-\frac{3}{2} \sin x+\frac{1}{2}}=1\) are a fascinating part of Mathematics for JEE Main. These equations involve trigonometric functions and often require careful analysis to solve. When dealing with trigonometric equations, it's essential to recognize that the solution may consist of multiple possibilities. This is because trigonometrical functions, such as sine and cosine, are periodic. Therefore, every solution tends to repeat after a certain interval. In the given exercise,

• The key is to find when the expression \(|\cos x|^{\sin^{2} x- \frac{3}{2} \sin x+\frac{1}{2}}=1\) holds true.
• Recall that any expression of the form \(a^b = 1\) is true if \(a=1\), \(a=-1\) with \(b\) being even, or \(b=0\).

Understanding these basic properties lends significant insight into solving complex trigonometric equations efficiently.

Quadratic Equations

Quadratic equations surface frequently, even in trigonometric contexts. The quadratic equation derived from the given problem is \(y^2 - \frac{3}{2}y + \frac{1}{2} = 0\), where \(y\) represents \(\sin x\).To solve this, we employ the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), for an equation of the form \( ax^2 + bx + c = 0 \). Here,

• \(a = 1\)
• \(b = -\frac{3}{2}\)
• \(c = \frac{1}{2}\)

Using these values, the solutions are \( y = 1 \) and \( y = \frac{1}{2} \). This gives potential values for \( \sin x \), which can then be used to determine \(x\). Understanding quadratic equations not only helps solve this step but also reinforces the concept of how algebraic manipulations can interconnect with trigonometry.

Trigonometrical Functions

In the realm of trigonometry, functions like sine and cosine are pivotal. These functions are periodic and defined across the unit circle. For cosine, \(\cos x = 1\) means \(x = 2n\pi\), and \(\cos x = -1\) corresponds to \((2n+1)\pi\). On the other hand, for sine, \(\sin x = 1\) occurs at \(x = (2n+1)\frac{\pi}{2}\), while \(\sin x = \frac{1}{2}\) happens at \(x = n\pi + (-1)^n\frac{\pi}{6}\).It’s essential to grasp these core behaviors to find the correct values of \(x\) in trigonometric contexts. By knowing these crucial points within the unit circle, you can efficiently solve trigonometric problems and accurately navigate through the maze that is trigonometric equations.Understanding trigonometrical functions deeply not only helps in mathematics but also enhances a learner's problem-solving abilities for exams like JEE Main.