Problem 78 - Algebra 2

Question

What are the solutions of \(2 \sin \theta-\sqrt{3}=0\) for \(0 \leq \theta<2 \pi ?\) F. \(\frac{\pi}{6}\) and \(\frac{5 \pi}{6}\) H. \(\frac{2 \pi}{3}\) and \(\frac{4 \pi}{3}\) G. \(\frac{\pi}{3}\) and \(\frac{2 \pi}{3}\) J. \(\frac{4 \pi}{3}\) and \(\frac{5 \pi}{3}\)

Step-by-Step Solution

Answer

F. \(\frac{\pi}{3}\) and \(\frac{5 \pi}{3}\)

Step 1: Isolate \(\sin(\theta)\)

First, to isolate \(\sin(\theta)\) start by adding \(\sqrt{3}\) to both sides of the equation. This gives us \(2 \sin \theta = \sqrt{3}\). Then, divide both sides by 2 to solve for \(\sin(\theta)\), resulting in \(\sin \theta = \frac{\sqrt{3}}{2}\).

Step 2: Find the values of \(\theta\)

Given that \(\sin \theta = \frac{\sqrt{3}}{2}\), and knowing that the sine function outputs this for the angles \(\frac{\pi}{3}\) and \(\frac{2\pi}{3}\) within the interval specified, one can conclude that \(\theta\) could be either of these two values. However, taking the periodicity of the sine function into account, since the sine is positive in the first and second quadrant, the possible values for \(\theta\) are actually \(\frac{\pi}{3}\), \(2\pi - \frac{\pi}{3} = \frac{5\pi}{3}\).

Key Concepts

Understanding the Sine Function

The sine function, denoted as \( \sin(\theta) \), is one of the primary functions in trigonometry and plays a crucial role in various fields like physics, engineering, and even music. At its core, the sine function relates an angle of a right triangle to the ratio of the length of the opposite side over the hypotenuse. However, in the unit circle approach, \( \theta \) represents an angle drawn from the positive x-axis in a counterclockwise direction, resulting in a point on the unit circle.
The sine value of this angle is simply the y-coordinate of this point. The range of the sine function is between -1 and 1, indicating the maximum and minimum values it can achieve. This makes it ideal for measuring smooth, periodic oscillations.

• The sine function is periodic, repeating its values every \(2\pi\) radians (360 degrees).
• This periodicity means that the function looks the same for every interval of \(2\pi\).
• The sine function has zeros at 0, \(\pi\), \(2\pi\), and so forth.

Understanding the sine function is fundamental for solving trigonometric equations effectively.

The Unit Circle and Its Role

Visualizing angles and their corresponding sine values gets much easier with the unit circle. The unit circle is a perfect circle with a radius of 1, centered at the origin of a coordinate plane. Every point on this circle can be defined by an angle \(\theta\), where \( \theta \) is measured in radians.
The unit circle helps in visualizing the sine as well as cosine and tangent functions. Here, the sine of an angle is depicted by the y-coordinate of the point on the circle where the terminal side of the angle intersects. Given the radius (r) is 1, it simplifies calculations, which is why it's so much favored in trigonometry.

• Allows for finding sine, cosine, and tangent of standard angles quickly.
• Facilitates the understanding of why certain angles have identical sine values (reflection over the y-axis).
• Useful for learning angle transformations and trigonometric identities.

The unit circle efficiently demonstrates why the values of \(\sin(\pi/3)\), which equals \(\sqrt{3}/2\), can also be mirrored in the second quadrant at \(\sin(2\pi/3)\). "Mirroring" refers to how an angle's sine value is the same as its "partner" angle in the other quadrant.

Finding Angle Solutions in Trigonometry

Solving trigonometric equations often involves finding possible angle solutions that satisfy the given equation. In this case, for the equation \(2\sin\theta - \sqrt{3} = 0\), once simplified to \(\sin\theta = \frac{\sqrt{3}}{2}\), the solutions can be derived using known properties of the sine function and the unit circle.
This involves identifying the angles known to produce the given sine value within a specified range, typically from 0 to \(2\pi\) in trigonometric contexts.
Here's how you can determine the right angle solutions:

• Identify the known angles that produce the required sine value, e.g., \(\sin\theta = \frac{\sqrt{3}}{2}\) is known for \(\theta = \pi/3\) and \(2\pi/3\).
• Consider the periodicity and symmetry properties of the sine function to find further solutions. The sine function's periodicity means that \(\sin\theta\) at \(\pi/3\) can also recurve at \(2\pi...\)
• Finally, observe the constraints of the problem, namely the interval \(0 \leq \theta < 2\pi\), which implies only certain possible solutions qualify.

Therefore, within the interval, \(\theta = \frac{\pi}{3} \) and \( \frac{5\pi}{3} \) make sense as valid angle solutions.