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

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

Correct answer: (C) \(n\pi + (-1)^n \frac{\pi}{6}, n \in I\).

Step 1: Understand the Equation

The given equation is \(|\cos x|^{\sin ^{2} x-\frac{3}{2} \sin x+\frac{1}{2}}=1\). For this expression to equal 1, the exponent must be 0 (since any number to the power of 0 is 1) or the base must be 1 or -1.

Step 2: Analyze the Base

The base is \(|\cos x|\), which equals 1 when \(\cos x = \pm 1\). This happens when \(x = n\pi\), where \(n\) is any integer.

Step 3: Analyze the Exponent

The exponent is \(\sin^2 x - \frac{3}{2} \sin x + \frac{1}{2} = 0\). Solving this quadratic in terms of \(\sin x\), we set \(a = \sin x\): \(a^2 - \frac{3}{2}a + \frac{1}{2} = 0\).

Step 4: Solve the Quadratic Equation

Using the quadratic formula \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), with \(b = -\frac{3}{2}\), \(a = 1\), and \(c = \frac{1}{2}\), calculate the roots: \(a = 1\) or \(a = \frac{1}{2}\). This means \(\sin x = 1\) or \(\sin x = \frac{1}{2}\).

Step 5: Identify Solutions for \(\sin x = 1\)

\(\sin x = 1\) when \(x = \frac{\pi}{2} + 2n\pi\). However, this is not always within the expressions considered in the answer choices, so other solutions must be considered.

Step 6: Identify Solutions for \(\sin x = \frac{1}{2}\)

\(\sin x = \frac{1}{2}\) when \(x = n\pi + (-1)^n \frac{\pi}{6}\), since one solution is in the first quadrant and another in the second within each period.

Step 7: Combine Possible Solutions

Combining \(x = n\pi\) from Step 2, \(x = \frac{\pi}{2} + 2n\pi\) (considered but not viable), and \(x = n\pi + (-1)^n \frac{\pi}{6}\) from Steps 5 and 6, we must select based on the matching answer choices.

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in algebra and arise frequently in trigonometric contexts. In our specific problem, the equation in the exponent involves terms of the sine function that are squared or linear. This forms a quadratic equation in terms of \(a = \sin x\). The standard form of a quadratic equation is \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants.
The solution involves finding values of \(\sin x\) by solving \(\sin^2 x - \frac{3}{2} \sin x + \frac{1}{2} = 0\). By substituting \(a = \sin x\), you apply the quadratic formula:

• Identify coefficients: \(b = -\frac{3}{2}\), \(a = 1\), \(c = \frac{1}{2}\).
• Calculate roots: \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• This results in \(a = 1\) or \(a = \frac{1}{2}\).

Understanding quadratic equations and how to extract real solutions has many applications. Not only does it help in problems like these but also in real life models where rates or areas might naturally form such curves.

Cosine Function

The cosine function, often encountered in right-angle triangles and waves, is critical in this problem. The base of our exponent involves the absolute value of the cosine function, \(|\cos x|\). Here, for the original expression to equal 1, managing the base when \(\cos x\) becomes \(\pm 1\) is fundamental:

• When \(|\cos x| = 1\), it implies \(\cos x = 1\) or \(\cos x = -1\).
• This occurs at specific angles \(x = n\pi\) (where \(n\) is an integer), meaning the function completes full cycles at multiples of \pi\.

The understanding of \(\cos x\) cycles is vital as it dictates energy wave forms, light, and sound waves. Its patterns highlight periodicity crucial in transforming and understanding wave functions across different mathematics and physics contexts.

Sine Function

The sine function varies between -1 and 1, influencing many natural rhythms and periodic phenomena, such as tides or sound waves. In our problem, the sine function dictates solutions for the quadratic in the exponent:

• The possible values for \(\sin x\) after solving the quadratic equation are \(1\) and \(\frac{1}{2}\).
• Solutions for \(\sin x = 1\) occur at \(x = \frac{\pi}{2} + 2n\pi\).
• For \(\sin x = \frac{1}{2}\), \(x = n\pi + (-1)^n \frac{\pi}{6}\).

Recognizing the \(\sin\) function’s properties, such as its half-wave at \(\frac{\pi}{2}\) or the first quadrant value at \(\frac{\pi}{6}\), provides insights not only into solving equations but also optimizing functionalities in trigonometry, allowing for precise modeling of cyclical behaviors.