Problem 8 - Precalculus with Calculus Previews

Question

Find all solutions of the given trigonometric equation if \(x\) represents a real number. $$ 2 \sin x=-1 $$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{7\pi}{6} + 2k\pi\) and \(x = \frac{11\pi}{6} + 2k\pi\), \(k \in \mathbb{Z}\).

Step 1: Isolate the Trigonometric Function

To solve the equation \(2 \sin x = -1\), first isolate \(\sin x\) by dividing both sides by 2. This gives us \(\sin x = -\frac{1}{2}\).

Step 2: Determine Reference Angles

We need to find the angles whose sine is \(-\frac{1}{2}\). Recall from the unit circle that \(\sin x = -\frac{1}{2}\) at \(x = \frac{7\pi}{6}\) and \(x = \frac{11\pi}{6}\) in the interval \([0, 2\pi)\).

Step 3: General Solution for the Sine Function

The general solution for \(\sin x = -\frac{1}{2}\) is given by \(x = \frac{7\pi}{6} + 2k\pi\) and \(x = \frac{11\pi}{6} + 2k\pi\), where \(k\) is any integer. This accounts for the periodic nature of the sine function and incorporates all possible solutions.

Key Concepts

Unit Circle

The unit circle is a crucial tool in trigonometry that helps us solve equations involving trigonometric functions. It is a circle with a radius of 1, centered at the origin of a coordinate plane. This simple shape allows us to understand the behavior of sine, cosine, and tangent functions based on the angle formed with the positive x-axis.
Each point on the unit circle corresponds to an angle \(x\), measuring the length of the arc from the positive x-axis.

• The x-coordinate of a point on the unit circle gives us the cosine of that angle.
• The y-coordinate gives us the sine.

These properties are why the unit circle is perfect for pinpointing exact values for sine and cosine at various angles. When solving equations like \(2 \sin x = -1\), we can use the unit circle to quickly identify angles where the sine (y-coordinate) is \(-\frac{1}{2}\). This corresponds to the exact angles \(\frac{7\pi}{6}\) and \(\frac{11\pi}{6}\) within a full rotation of \[0, 2\pi\].
Understanding the unit circle aids in visualizing how these functions behave over different intervals and how the angles repeat, making it essential for determining trigonometric equations' solutions.

General Solution

Finding the general solution of a trigonometric equation means identifying all the possible values for the variable that satisfy it. Trigonometric functions like sine and cosine are periodic, meaning they repeat their values at regular intervals. For sine and cosine, this interval is \(2\pi\).
For the equation \(\sin x = -\frac{1}{2}\), we determined specific angles from the unit circle that solve the equation within the \[0, 2\pi\] range: \(\frac{7\pi}{6}\) and \(\frac{11\pi}{6}\). However, these angles recur every full rotation of the circle. To capture all solutions:

• We add \(2k\pi\) to each of the specific solutions, where \(k\) is any integer.
• This accounts for the periodic nature of trigonometric functions.

Thus, the general solution becomes \(x = \frac{7\pi}{6} + 2k\pi\) and \(x = \frac{11\pi}{6} + 2k\pi\). Each integer value of \(k\) results in a new solution by moving forward or backward around the circle, encompassing all possible angles satisfying the original equation.

Reference Angles

Reference angles play an invaluable role in solving trigonometric equations, especially when working with negative or non-standard angles. A reference angle is the acute angle formed by the terminal side of an angle and the x-axis.
To solve \(\sin x = -\frac{1}{2}\) using reference angles, we look for angles with a known sine value given by the unit circle. Since \(-\frac{1}{2}\) is negative, we focus on angles in the third and fourth quadrants because that's where the sine is negative.
From the unit circle, we know that:

• \(\sin \left(\frac{\pi}{6}\right)\) equals \(\frac{1}{2}\) in the first quadrant.
• Thus, the reference angles for \(-\frac{1}{2}\) are \(\frac{7\pi}{6}\) and \(\frac{11\pi}{6}\), reflecting similar arrangements in the third and fourth quadrants respectively.

By using the symmetry of the circle and knowledge of these specific angles, we leverage reference angles to find solutions efficiently across different quadrants, ensuring understanding of how these angles conform to trigonometric functions' properties.