Chapter 8: Problem 3
Find all solutions of the equation. $$2 \sin x-1=0$$
Short Answer
Expert verified
The solutions are \(x = \frac{\pi}{6} + 2k\pi\) and \(x = \frac{5\pi}{6} + 2k\pi\) for integer \(k\).
Step by step solution
01
Isolate the Sine Function
The original equation is \(2 \sin x - 1 = 0\). To isolate \(\sin x\), add 1 to both sides, giving \(2 \sin x = 1\). Next, divide both sides by 2, resulting in \(\sin x = \frac{1}{2}\).
02
Identify General Solutions for Sine
We know that \(\sin x = \frac{1}{2}\) occurs at specific angles. The general solutions for \(\sin x = \frac{1}{2}\) are \(x = \frac{\pi}{6} + 2k\pi\) and \(x = \frac{5\pi}{6} + 2k\pi\) where \(k\) is any integer. These angles correspond to the two quadrants where sine is positive.
03
Write the Solutions
The complete solution set is \(x = \frac{\pi}{6} + 2k\pi\) and \(x = \frac{5\pi}{6} + 2k\pi\) for \(k \in \mathbb{Z}\). These sequences represent all the angular positions where \(\sin x = \frac{1}{2}\) on a unit circle, covering rotations of full circles.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sine Function
The sine function is a fundamental concept in trigonometry, describing the ratio of the length of the opposite side to the hypotenuse in a right triangle. In mathematical terms, for an angle \( x \) in a right-angled triangle, \( \sin x \) gives this ratio. This function is periodic with a period of \( 2\pi \), meaning it repeats its pattern every \( 2\pi \) radians. The range of the sine function is from -1 to 1, which means it cannot produce values outside this interval. In many problems, especially trigonometric equations, you'll encounter expressions like \( 2 \sin x - 1 = 0 \), where isolating \( \sin x \) helps solve the equation. Here's how:
- Firstly, solve for \( \sin x \) by using algebraic transformations, such as addition or division.
- Once \( \sin x \) is isolated, its value is set within the permissible range of -1 to 1.
Angle Solutions
Finding angle solutions involves determining all possible angles \( x \) that satisfy a given trigonometric equation. For example, if \( \sin x = \frac{1}{2} \), we need to find all angles \( x \) for which this value is true. The sine function equals \( \frac{1}{2} \) at specific points within one full cycle (0 to \( 2\pi \)). These angles for \( \sin x = \frac{1}{2} \) are:
- \( x = \frac{\pi}{6} \)
- \( x = \frac{5\pi}{6} \)
- The general solutions include multiples of the full cycle available through \( 2k\pi \), where \( k \) is any integer.
- This is represented mathematically as \( x = \frac{\pi}{6} + 2k\pi \) and \( x = \frac{5\pi}{6} + 2k\pi \), encompassing all possible angles satisfying the equation.
Unit Circle
The unit circle is an essential tool in trigonometry, representing all angles and their corresponding sine and cosine values on a circle with a radius of 1. On this circle:
- The x-coordinate represents \( \cos x \).
- The y-coordinate represents \( \sin x \).
- Find points where the y-coordinate is \( \frac{1}{2} \).
- These points on the unit circle correspond to angles \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \).
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that hold true for all angles. These identities are essential in simplifying trigonometric expressions and solving equations. Some key purposes include:
- Transforming complex equations to simpler forms.
- Relating different trigonometric functions to each other, such as \( \sin^2 x + \cos^2 x = 1 \).
- Identities like the Pythagorean identity can help verify solutions or check consistency across methods.
- Understanding symmetries and periodicities of these functions simplify finding general solutions.