Solving 2sin(θ) + 1 = 0 Find Solutions In The Interval 0 ≤ Θ < 2π

Jul 9, 2025 by Admin 66 views

Solving Trigonometric Equations: Finding Solutions for 2sin(θ) + 1 = 0 in the Interval 0 ≤ θ < 2π

Introduction

In the realm of trigonometry, solving equations is a fundamental skill. These equations often involve trigonometric functions such as sine, cosine, and tangent. Understanding how to solve these equations is crucial for various applications in physics, engineering, and other scientific fields. In this article, we will delve into the process of solving a trigonometric equation within a specified interval. Specifically, we will focus on finding the solutions to the equation 2sin(θ) + 1 = 0 within the interval 0 ≤ θ < 2π. This interval represents one full revolution around the unit circle, and our goal is to identify all angles θ within this range that satisfy the given equation.

Trigonometric equations, like algebraic equations, require us to isolate the variable. However, in trigonometric equations, the variable is typically an angle, and it is embedded within a trigonometric function. To solve these equations, we will utilize our knowledge of trigonometric identities, the unit circle, and the properties of trigonometric functions. The unit circle, a circle with a radius of 1 centered at the origin, provides a visual representation of the values of sine, cosine, and tangent for different angles. By understanding the unit circle, we can readily identify angles that correspond to specific trigonometric values.

This article will guide you through the step-by-step process of solving the equation 2sin(θ) + 1 = 0 in the interval 0 ≤ θ < 2π. We will begin by isolating the sine function, then use the unit circle to find the angles that satisfy the equation. Finally, we will express the solutions within the given interval. By the end of this article, you will have a solid understanding of how to approach and solve trigonometric equations of this type.

Isolating the Sine Function

To begin solving the trigonometric equation 2sin(θ) + 1 = 0, our primary goal is to isolate the sine function. This involves algebraic manipulation to get the sin(θ) term by itself on one side of the equation. The process is similar to solving linear equations in algebra, where we aim to isolate the variable. In this case, our "variable" is the sine of the angle θ.

The first step in isolating sin(θ) is to subtract 1 from both sides of the equation. This operation maintains the equality of the equation while moving the constant term to the right side. The equation then becomes:

2sin(θ) = -1

Next, we need to get sin(θ) completely by itself. To do this, we divide both sides of the equation by 2. This operation isolates sin(θ) on the left side and gives us the value it must equal on the right side. The equation now looks like this:

sin(θ) = -1/2

At this point, we have successfully isolated the sine function. We now know that we are looking for angles θ whose sine is equal to -1/2. This is a crucial step because it allows us to focus on the specific values of the sine function and relate them to angles on the unit circle. The next step involves using our knowledge of the unit circle to identify the angles that satisfy this condition. Understanding how to isolate trigonometric functions is a fundamental skill in solving trigonometric equations, and this process is applicable to a wide range of problems.

Utilizing the Unit Circle

Now that we have isolated the sine function and determined that sin(θ) = -1/2, the next crucial step is to utilize the unit circle to find the angles θ that satisfy this condition. The unit circle is an invaluable tool for solving trigonometric equations because it provides a visual representation of the sine, cosine, and tangent values for all angles between 0 and 2π.

Recall that on the unit circle, the sine of an angle θ corresponds to the y-coordinate of the point where the terminal side of the angle intersects the circle. Therefore, we are looking for points on the unit circle where the y-coordinate is -1/2. To identify these points, visualize the unit circle and consider where the y-coordinate would be negative. Since the y-coordinate is negative in the third and fourth quadrants, we know that our solutions will lie in these quadrants.

Next, we need to recall the reference angles associated with a sine value of 1/2. The reference angle is the acute angle formed between the terminal side of the angle and the x-axis. We know that sin(π/6) = 1/2, so π/6 is our reference angle. Now, we need to find the angles in the third and fourth quadrants that have a reference angle of π/6.

In the third quadrant, the angle is given by π + π/6 = 7π/6. This angle is π radians (180 degrees) plus the reference angle, placing it in the third quadrant where both x and y coordinates are negative. In the fourth quadrant, the angle is given by 2π - π/6 = 11π/6. This angle is 2π radians (360 degrees) minus the reference angle, placing it in the fourth quadrant where the x-coordinate is positive and the y-coordinate is negative.

Therefore, the angles 7π/6 and 11π/6 are the solutions to the equation sin(θ) = -1/2 within the interval 0 ≤ θ < 2π. These angles correspond to the points on the unit circle where the y-coordinate is -1/2. The unit circle provides a clear and intuitive way to visualize and determine the solutions to trigonometric equations.

Identifying Solutions in the Interval 0 ≤ θ < 2π

Having identified the angles 7π/6 and 11π/6 as potential solutions to the equation 2sin(θ) + 1 = 0, we must now confirm that these solutions fall within the specified interval of 0 ≤ θ < 2π. This interval represents one complete revolution around the unit circle, starting from 0 radians and ending just before 2π radians. It is crucial to ensure that our solutions lie within this range to satisfy the original problem's conditions.

The angle 7π/6 is greater than π (which is equivalent to 6π/6) and less than 2π. Specifically, 7π/6 is located in the third quadrant of the unit circle, as we discussed earlier. Therefore, 7π/6 clearly falls within the interval 0 ≤ θ < 2π.

Similarly, the angle 11π/6 is greater than 3π/2 (which is equivalent to 9π/6) and less than 2π (which is equivalent to 12π/6). This angle is located in the fourth quadrant of the unit circle, where the sine function is negative. Thus, 11π/6 also falls within the interval 0 ≤ θ < 2π.

Since both 7π/6 and 11π/6 lie within the interval 0 ≤ θ < 2π, we can confidently conclude that these are the solutions to the equation 2sin(θ) + 1 = 0 within the given interval. There are no other angles within this range that satisfy the equation, as we have considered all possible quadrants and reference angles using the unit circle.

Expressing the Solutions

Having determined that the solutions to the equation 2sin(θ) + 1 = 0 within the interval 0 ≤ θ < 2π are 7π/6 and 11π/6, it is important to express these solutions clearly and concisely. The standard way to present the solutions is to list them in a set, typically enclosed in curly braces. This notation clearly indicates that we are providing a set of values that satisfy the given equation within the specified interval.

Therefore, the solutions to the equation 2sin(θ) + 1 = 0 in the interval 0 ≤ θ < 2π can be expressed as:

{7π/6, 11π/6}

This set notation provides a clear and unambiguous way to communicate the solutions. It is important to include the curly braces to indicate that we are referring to a set of values rather than individual values. This notation is widely used in mathematics to represent sets, and it is essential for conveying the solutions accurately.

In summary, the solutions to the trigonometric equation 2sin(θ) + 1 = 0 within the interval 0 ≤ θ < 2π are 7π/6 and 11π/6, and these solutions are best expressed as the set {7π/6, 11π/6}. This final step of expressing the solutions ensures that the answer is presented in a clear and professional manner.

Conclusion

In this article, we have thoroughly explored the process of solving the trigonometric equation 2sin(θ) + 1 = 0 within the interval 0 ≤ θ < 2π. We began by understanding the importance of solving trigonometric equations and the role they play in various scientific and engineering applications. We then systematically worked through the steps required to find the solutions.

First, we focused on isolating the sine function. This involved algebraic manipulation to get the sin(θ) term by itself on one side of the equation, resulting in the simplified equation sin(θ) = -1/2. This step is crucial because it allows us to focus on the specific values of the sine function and relate them to angles on the unit circle.

Next, we utilized the unit circle to identify the angles that satisfy the equation sin(θ) = -1/2. By understanding that the sine of an angle corresponds to the y-coordinate on the unit circle, we were able to visualize and determine the angles 7π/6 and 11π/6 as potential solutions. The unit circle serves as a powerful tool for solving trigonometric equations, providing a visual representation of trigonometric values for different angles.

We then verified that these potential solutions, 7π/6 and 11π/6, fell within the specified interval of 0 ≤ θ < 2π. This step is essential to ensure that our solutions meet the conditions of the original problem. Since both angles lie within the interval, we confirmed that they are indeed valid solutions.

Finally, we expressed the solutions clearly and concisely as the set {7π/6, 11π/6}. This notation is the standard way to represent a set of solutions in mathematics, and it provides a clear and unambiguous answer to the problem.

By following these steps, we have successfully solved the trigonometric equation 2sin(θ) + 1 = 0 within the interval 0 ≤ θ < 2π. This process demonstrates the importance of algebraic manipulation, the unit circle, and careful consideration of intervals when solving trigonometric equations. The skills and techniques learned in this article can be applied to a wide range of trigonometric problems, making this a valuable exercise for anyone studying trigonometry or related fields.

Practice Problems

To further solidify your understanding of solving trigonometric equations, try working through these practice problems. Each problem involves solving a trigonometric equation within a specified interval. Remember to follow the steps outlined in this article: isolate the trigonometric function, use the unit circle to find potential solutions, verify that the solutions fall within the interval, and express the solutions clearly.

Solve the equation 2cos(θ) - 1 = 0 for 0 ≤ θ < 2π.
Find all solutions to the equation √3 tan(θ) + 1 = 0 in the interval 0 ≤ θ < 2π.
Determine the solutions for 2sin(θ) - √3 = 0 where 0 ≤ θ < 2π.
Solve the equation cos(2θ) = 1/2 for 0 ≤ θ < 2π (Hint: This involves a double angle).
Find the solutions to sin(θ) + 1 = 0 in the interval 0 ≤ θ < 2π.

Working through these problems will help you develop confidence and proficiency in solving trigonometric equations. Remember to utilize the unit circle and your knowledge of trigonometric identities to find the solutions. Good luck!