Problem 3 - Algebra and Trigonometry

Question

Find all solutions of the equation. $$2 \sin x-1=0$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{\pi}{6} + 2k\pi\) and \(x = \frac{5\pi}{6} + 2k\pi\) for integer \(k\).

Step 1: 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 \(\sin x = \frac{1}{2}\).

Step 2: 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.

Step 3: 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.

Key Concepts

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.

Understanding the sine function's limitations and behaviors is key to tackling trigonometric equations effectively.

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:

• \( x = \frac{\pi}{6} \)
• \( x = \frac{5\pi}{6} \)

Beyond these points, trigonometric solutions must consider the periodic nature of the sine function. This means:

• 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.

Understanding how to find these angle solutions is crucial for solving trigonometric equations.

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 \( \sin x \).

Because the circle's radius is 1, any point on the circle has coordinates \( (\cos x, \sin x) \), indicating both cosine and sine values of angle \( x \). When solving \( \sin x = \frac{1}{2} \), you can locate points on the unit circle that produce this value. Here's how:

• Find points where the y-coordinate is \( \frac{1}{2} \).
• These points on the unit circle correspond to angles \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \).

The unit circle helps visualize why these angles are solutions, as sine reflects angles above the x-axis (positive y-values) within certain quadrants. This visual representation supports solving equations and understanding angle relationships.

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 \( \sin^2 x + \cos^2 x = 1 \).

In solving \( \sin x = \frac{1}{2} \), identities can provide insights and alternative routes to discovery. For example:

• 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.

Mastering trigonometric identities enables deeper comprehension of trigonometric equations, providing a solid foundation for tackling complex mathematical problems.