Problem 18
Question
Find all solutions of the equation in the interval \([0,2 \pi)\). $$\sin x=-\frac{1}{2}$$
Step-by-Step Solution
Verified Answer
The solutions to the equation \( \sin x = -\frac{1}{2} \) in the interval \([0,2\pi)\) are \( x= \frac{7\pi}{6} \) and \( x= \frac{11\pi}{6} \)
1Step 1: Identify the Basic Solution
Firstly, we identify the principal or basic solution to the equation. From the knowledge of special angles in trigonometry, we know that \( \sin x = -\frac{1}{2} \) for \( x= \frac{7\pi}{6} \) and \( x= \frac{11\pi}{6} \).
2Step 2: Use the Periodic Property of Sine Function
Next we use the periodic property of the sine function to find all the other solutions in the given interval. The sine function has a periodicity of \(2\pi\), which means \( \sin(x + 2n\pi) = \sin x \), where n is any integer, which provides the remaining solutions in the specified interval. In this case, however, since our interval is between 0 and \(2\pi\), n can only be 0, because for any other integer value the results would fall outside the given interval.
3Step 3: List All Solutions
Finally, we can list all solutions within the interval \( [0,2\pi) \). There are two solutions for the original equation within the given interval, and they are \( x= \frac{7\pi}{6} \) and \( x= \frac{11\pi}{6} \)
Key Concepts
Special AnglesSine FunctionPeriodicity
Special Angles
When solving trigonometric equations, the concept of special angles is vital. Special angles are specific angles with well-known trigonometric values. These angles, found in the unit circle, include common degrees such as 30°, 45°, 60°, and their multiples. In radians, these angles correspond to \( \frac{\pi}{6} \), \( \frac{\pi}{4} \), \( \frac{\pi}{3} \), and so on.
Identifying these angles helps in solving trigonometric equations quickly because their sine, cosine, and tangent values are easy to remember. For instance, \( \sin \left( \frac{\pi}{6} \right) = \frac{1}{2} \), which is a commonly used value.
In problems like "\( \sin x = -\frac{1}{2} \)", knowing that \( \sin \left( \frac{\pi}{6} \right) = \frac{1}{2} \) allows us to determine that an angle where sine is \(-\frac{1}{2}\) must lie in the third or fourth quadrant. This reasoning helps find basic solutions, such as \( x = \frac{7\pi}{6} \) and \( x = \frac{11\pi}{6} \), which are multiples of special angles.
Identifying these angles helps in solving trigonometric equations quickly because their sine, cosine, and tangent values are easy to remember. For instance, \( \sin \left( \frac{\pi}{6} \right) = \frac{1}{2} \), which is a commonly used value.
In problems like "\( \sin x = -\frac{1}{2} \)", knowing that \( \sin \left( \frac{\pi}{6} \right) = \frac{1}{2} \) allows us to determine that an angle where sine is \(-\frac{1}{2}\) must lie in the third or fourth quadrant. This reasoning helps find basic solutions, such as \( x = \frac{7\pi}{6} \) and \( x = \frac{11\pi}{6} \), which are multiples of special angles.
Sine Function
The sine function is a fundamental trigonometric function used to analyze periodic phenomena. It takes an angle as input and returns the ratio between the length of the opposite side to the hypotenuse in a right-angled triangle.
Mathematically, the range of the sine function is from \(-1\) to \(1\), which means its values oscillate between these limits. Let's summarize some important features:
This is especially useful in solving equations like \(\sin x = -\frac{1}{2}\), where it indicates the answers are not just positive or zero but can also be negative.
Mathematically, the range of the sine function is from \(-1\) to \(1\), which means its values oscillate between these limits. Let's summarize some important features:
- At \(x = 0\), \(\sin x = 0\).
- At \(x = \frac{\pi}{2}\), \(\sin x = 1\).
- At \(x = \pi\), \(\sin x = 0\).
- At \(x = \frac{3\pi}{2}\), \(\sin x = -1\).
This is especially useful in solving equations like \(\sin x = -\frac{1}{2}\), where it indicates the answers are not just positive or zero but can also be negative.
Periodicity
Periodicity refers to the repeating nature of a function. For sine and cosine functions, this period is \(2\pi\). This means that after an interval of \(2\pi\), the sine values repeat.
For instance, if \(\sin x = a\), then \(\sin (x + 2n\pi) = a\), where \(n\) is any integer. This property of periodicity is particularly practical when we want to find all solutions of a trigonometric equation within a specific interval, such as \([0, 2\pi)\).
In the equation \(\sin x = -\frac{1}{2}\), by using periodicity, we understand that moving \(2\pi\) units in either direction keeps the sine value constant. However, since the interval \([0, 2\pi)\) limits our scope to just one full cycle, we restrict \(n\) to 0 to ensure our solutions are within these bounds.
For instance, if \(\sin x = a\), then \(\sin (x + 2n\pi) = a\), where \(n\) is any integer. This property of periodicity is particularly practical when we want to find all solutions of a trigonometric equation within a specific interval, such as \([0, 2\pi)\).
In the equation \(\sin x = -\frac{1}{2}\), by using periodicity, we understand that moving \(2\pi\) units in either direction keeps the sine value constant. However, since the interval \([0, 2\pi)\) limits our scope to just one full cycle, we restrict \(n\) to 0 to ensure our solutions are within these bounds.
- This results in solutions like \( x = \frac{7\pi}{6} \)
- and \( x = \frac{11\pi}{6} \)
Other exercises in this chapter
Problem 18
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Use a graphing utility to approximate the solutions of the equation in the interval \([0,2 \pi) .\) If possible, find the exact solutions algebraically. $$\tan
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