Problem 155 - Calculus Volume 1

Question

Solve the trigonometric equations on the interval \(0 \leq \theta<2 \pi\). \(2 \sin \theta-1=0\)

Step-by-Step Solution

Answer

\(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\)

Step 1: Isolate the Trigonometric Function

Start with the equation \(2 \sin \theta - 1 = 0\). To isolate \( \sin \theta \), add 1 to both sides: \(2 \sin \theta = 1\). Then divide both sides by 2 to solve for \( \sin \theta \), yielding \( \sin \theta = \frac{1}{2}\).

Step 2: Determine the General Solutions for Sine

The solutions for \( \sin \theta = \frac{1}{2} \) are angles where the sine of the angle is \( \frac{1}{2} \). From the unit circle or sine chart, these are \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \). These solutions are within one full circle of the unit circle.

Step 3: Verify the Solutions Within the Given Interval

The interval provided is \( 0 \leq \theta < 2\pi \). The solutions \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \) are both within this range. Therefore, these solutions satisfy the constraints of the problem.

Key Concepts

Unit Circle

The unit circle is a fundamental tool in trigonometry. It's essentially a circle with a radius of 1 unit. This circle is important because it helps us understand the relationships between different trigonometric functions and angles.

If you imagine this circle centered at the origin of a coordinate plane, each point on the circle corresponds to a specific angle, measured in radians, starting from the positive x-axis. This angle is often named as \( \theta \). When working with the unit circle, the key thing to remember is:

• The x-coordinate of a point on the unit circle represents the cosine of the angle \( \theta \).
• The y-coordinate represents the sine of \( \theta \).

For the angle \( \theta = \frac{\pi}{6} \), its sine (which is the y-coordinate on the unit circle) is \( \frac{1}{2} \). Similarly, at \( \theta = \frac{5\pi}{6} \), the sine value is also \( \frac{1}{2} \). Understanding this circle and the positioning of angles on it is crucial for solving trigonometric equations efficiently.

Sine Function

The sine function is a trigonometric function that plays a crucial role in mathematics. It's defined in terms of a right triangle or a unit circle. In the context of the unit circle:

• The sine of an angle \( \theta \) is the y-coordinate of the corresponding point on the circle.

This function repeats its values in a predictable cycle, as it completes a full circle of \( 2\pi \) radians. For the sine function, this repetition is called periodicity. Its typical form looks like this:

\[ y = \sin(\theta) \]

In our problem, we set \( \sin(\theta) = \frac{1}{2} \). This equation tells us to find all angles \( \theta \) where the sine value equals \( \frac{1}{2} \).

The sine function's understanding can be visualized easier on the unit circle, where specific angles like \( \theta = \frac{\pi}{6} \) provide a sine of \( \frac{1}{2} \). This cyclic property of sine is key to finding solutions quickly.

Interval Solutions

When solving trigonometric equations, identifying solutions within a specific interval is key. Intervals, such as \( 0 \leq \theta < 2\pi \), determine the range within which angles are valid.

An interval like this represents one complete rotation around the unit circle. It's important to check if your solutions fall within this interval.

• For example, in our problem, \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \) are valid solutions because both are between \( 0 \) and \( 2\pi \).

When working with intervals, keep these tips in mind:

• Pay attention to the interval's boundaries. Here, \( 0 \) is included, while \( 2\pi \) is not.
• Multiple solutions might exist within the interval for periodic functions like sine.

Being adept at interval plotting and cross-checking ensures that all potential solutions are considered and validated.