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Magazine Chapter 7: Problem 241
For the following exercises, solve exactly on \([0,2 \pi)\) $$ 2 \sin (3 \theta)=1 $$
Short Answer
Expert verified
The solutions are \(\theta = \frac{\pi}{18}, \frac{13\pi}{18}, \frac{25\pi}{18}, \frac{5\pi}{18}, \frac{17\pi}{18}, \frac{29\pi}{18}\).
Step by step solution
01
Isolate the Trigonometric Function
Start by isolating the trigonometric function. Given the equation \( 2 \sin(3\theta) = 1 \), divide both sides by 2 to get: \[ \sin(3\theta) = \frac{1}{2} \]
02
Find the Principal Values
Determine the values of \(3\theta\) for which \(\sin(3\theta) = \frac{1}{2}\) within the unit circle. We know that \(\sin\theta = \frac{1}{2}\) at \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\), so:\[ 3\theta = \frac{\pi}{6} + 2k\pi \text{ and } 3\theta = \frac{5\pi}{6} + 2k\pi \text{ for } k \in \mathbb{Z} \]
03
Solve for \( \theta \)
Solve for \(\theta\) by dividing each term by 3. This gives:\[ \theta = \frac{\pi}{18} + \frac{2k\pi}{3} \text{ and } \theta = \frac{5\pi}{18} + \frac{2k\pi}{3} \]
04
Find Solutions in the Interval \([0, 2\pi)\)
Find all values of \(\theta\) that are within the specified interval \([0, 2\pi)\). Starting with \(\theta = \frac{\pi}{18} + \frac{2k\pi}{3}\):- For \(k = 0\), \(\theta = \frac{\pi}{18}\)- For \(k = 1\), \(\theta = \frac{13\pi}{18}\)- For \(k = 2\), \(\theta = \frac{25\pi}{18}\) (which is still less than \(2\pi\))Next, for \(\theta = \frac{5\pi}{18} + \frac{2k\pi}{3}\):- For \(k = 0\), \(\theta = \frac{5\pi}{18}\)- For \(k = 1\), \(\theta = \frac{17\pi}{18}\)- For \(k = 2\), \(\theta = \frac{29\pi}{18}\) (which is still less than \(2\pi\))The solutions are \(\theta = \frac{\pi}{18}, \frac{13\pi}{18}, \frac{25\pi}{18}, \frac{5\pi}{18}, \frac{17\pi}{18}, \frac{29\pi}{18}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Unit Circle
The unit circle is a powerful tool used frequently in trigonometry. It is a circle with a radius of 1 centered at the origin of the coordinate plane. By using the unit circle, we can easily find the sine, cosine, and tangent values of various angles.
- Coordinates on the unit circle, such as ( \( \frac{\sqrt{3}}{2} \) , \(\frac{1}{2}\) ), indicate the cosine and sine values of the angle corresponding to that point.
- The angle is measured counterclockwise from the positive x-axis.
Sin Function
The sine function is one of the fundamental trigonometric functions. It takes an angle as input and returns a value between -1 and 1. The sine value of an angle corresponds to the y-coordinate of the point on the unit circle.
- For example, \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \) because the y-coordinate of the angle \(\frac{\pi}{6}\) is \( \frac{1}{2} \).
- The sine function is periodic with a period of \(2\pi\) , meaning the pattern repeats every \(2\pi\) radians.
- It is also an odd function: \( \sin(-\theta) = -\sin(\theta) \).
Angle Solutions
Angle solutions in trigonometric equations can have multiple values due to the periodic nature of trigonometric functions. When solving an equation like \(\sin(3\theta) = \frac{1}{2}\), we need to find all possible angles that satisfy the equation.
- The primary angles for which \(\sin(\theta) = \frac{1}{2}\) are \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\) according to the unit circle.
- Incorporating the multiplier 3 in \(3\theta\), each solution branches into intervals of \(2\pi\) full rotations, creating a series of angle solutions.
Interval [0, 2蟺)
The interval \([0, 2\pi)\) is a common range to express angles in trigonometry. This interval includes angles starting from 0, just up to but not including \(2\pi\).
- The notation \([0, 2\pi)\) indicates a closed interval at 0 but open (not including) at \(2\pi\).
- Within this interval, each angle represents a unique position on the unit circle.
- In solving our equation, checking each calculated \(\theta\) to fit within \([0, 2\pi)\) ensures the solutions are valid within a single rotation around the circle.
