Problem 11 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Solve for \(x\) : $$ \sin 2 x<\frac{1}{2} $$

Step-by-Step Solution

Answer

The solution to the inequality is \(x \in (0 + 2n\pi, 5\pi/6 + 2n\pi) \cup (\pi + 2n\pi, 2\pi + 2n\pi), n \in \mathbb{Z}\)

Step 1: Understanding the Sine Function

The sine function oscillates between -1 and 1. For \(0 \leq x \leq \pi\), sine function is less than or equal to 1/2 at \(\pi/6\) and \(5\pi/6\). Therefore the inequality \(\sin x < 1/2\) has the solutions \(x < \pi/6\) and \(\pi < X < 5\pi/6\) in the interval \([0, 2\pi]\). We'll use this property later.

Step 2: Apply the Double Angle Formula

The double angle formula states that \(\sin 2x = 2\sin x \cos x\). Our inequality becomes \(2\sin x \cos x < 1/2\). We can simplify this inequality to \(4\sin x \cos x < 1\). This is the same as \(2\sin x < 1 / 2\cos x\). Therefore, \(\sin x < 1 / 4\cos x\). This inequality has solutions where \(\cos x\) is positive and \(\sin x < 1 / 4\). The inequality is true for \(0 < x < \pi/6\) and \(\pi < x < 5\pi/6\). In the range where \(\cos x\) is negative, we have to flip the inequality sign and find \(\sin x > 1 / 4\cos x\). The inequality is true for \(\pi/6 < x < \pi\) and \(5\pi/6 < x < 2\pi\).

Step 3: Find the Final Solution Set

Combining the solution sets in the previous step, we have possible solutions in the following ranges: \(0 < x < \pi/6\), \(\pi < x < 5\pi/6\), \(\pi/6 < x < \pi\) and \(5\pi/6 < x < 2\pi\). We can express these intervals in a more compact way as \(0 < x < 5\pi/6\) and \(\pi < x < 2\pi\). Repeating these solutions for every \(2\pi\) interval, the final solution set for this inequality is \(x \in (0 + 2n\pi, 5\pi/6 + 2n\pi) \cup (\pi + 2n\pi, 2\pi + 2n\pi), n \in \mathbb{Z}\).

Key Concepts

Understanding the Sine Function

The sine function, denoted as \( \sin x \), is a fundamental trigonometric function that maps angles to the ratio of the opposite side over the hypotenuse in a right triangle. Its values range between \(-1\) and \(1\), creating a smooth, wave-like curve known as a sine wave.

For angles from \(0\) to \(2\pi\), the sine function achieves a maximum of \(1\) and a minimum of \(-1\). Specifically, it is less than or equal to \(1/2\) at \(\pi/6\) and \(5\pi/6\). This cyclical pattern means the function repeats its values over intervals of \(2\pi\). Knowing where \( \sin x \) attains certain values helps in solving inequalities like \( \sin x < 1/2 \).

• At \(x = \pi/6\), \( \sin x = 1/2\).
• At \(x = \pi\), \( \sin x = 0\).
• For \( \sin x < 1/2\), solutions are found between \(0 < x < \pi/6\) and \(\pi < x < 5\pi/6\).

Exploring the Double Angle Formula

The double angle formula is a powerful tool in trigonometry, especially for simplifying expressions like \( \sin 2x \). According to this formula, \( \sin 2x = 2 \sin x \cos x \). This transforms our inequality \( \sin 2x < 1/2 \) into \( 2 \sin x \cos x < 1/2 \).

By simplifying further, we get \( 4 \sin x \cos x < 1 \), or equivalently, \( \sin x < 1/4 \cos x \). This helps break down more complex trigonometric equations into manageable parts.

• Divide both sides by \(2\): \( \sin x \cos x < 1/4\).
• Ensure \(\cos x\) is positive: \( \sin x < 1/4 \cos x\).
• Find intervals like \(0 < x < \pi/6\) and \(\pi < x < 5\pi/6\) for positive \(\cos x\).

This operation reveals conditions where one trigonometric function is dominant over another, providing solutions to inequalities.

Solving Inequalities in Trigonometry

Trigonometric inequalities like \( \sin 2x < 1/2 \) require an understanding of how different sides of the function influence solutions. Inequalities can illustrate where a particular function value lies in relation to a given constant.

Consider the various intervals where sine and cosine functions behave differently due to their periodic nature. For this exercise, solutions were found in segments: \(0 < x < 5\pi/6\) and \(\pi < x < 2\pi\).

• Identify intervals of increasing or decreasing behavior.
• Evaluate signs of \(\cos x\) across intervals to determine inequality solutions.
• Combine overlapping or adjacent intervals for comprehensive solutions.

These approaches allow us to sketch solution sets vividly, covering all \(x\) solutions in successive cycles of \(2\pi\), i.e., repeating the solution in every \(2\pi\) interval.