Solving -2sin(θ) = -√2 Exact Solutions On The Interval 0 ≤ Θ < 2π

In the realm of trigonometry, solving equations is a fundamental skill. When seeking exact solutions within a specific interval, such as $0 \\\leq \\theta < 2\\\[Pi]$, we delve into the intricacies of trigonometric functions and their periodic nature. This article provides a comprehensive guide to solving the trigonometric equation $-2sin(θ) = -\\sqrt{2}$ within the specified interval, offering step-by-step explanations and insights into the underlying concepts. Understanding trigonometric equations is crucial for various applications in physics, engineering, and mathematics, making this a valuable skill for students and professionals alike. The ability to find exact solutions is particularly important, as it allows for precise calculations and avoids the rounding errors that can occur with approximate solutions. By mastering these techniques, you will be well-equipped to tackle a wide range of trigonometric problems.

Understanding the Problem

Before diving into the solution, let's first understand the given problem. We are tasked with finding all exact solutions for the equation $-2sin(θ) = -\\sqrt{2}$ within the interval $0 \\\leq \\\theta < 2\\\pi$. This means we need to identify all angles θ within one full rotation (0 to 360 degrees or 0 to $2\\\[Pi]$ radians) that satisfy the equation. Trigonometric equations often have multiple solutions due to the periodic nature of trigonometric functions. The sine function, for instance, repeats its values every $2\\\[Pi]$ radians. Therefore, it's crucial to consider all possible solutions within the given interval. Furthermore, the equation involves a negative coefficient, which affects the sign of the sine function. Recall that sine is positive in the first and second quadrants and negative in the third and fourth quadrants. This knowledge will be essential in identifying the correct quadrants for our solutions. By carefully considering these aspects, we can approach the problem systematically and ensure that we find all the exact solutions within the specified interval.

Isolating the Trigonometric Function

To begin solving the equation, our initial step involves isolating the sine function. This is achieved by dividing both sides of the equation $-2sin(θ) = -\\sqrt{2}$ by -2. This algebraic manipulation simplifies the equation and allows us to focus on the sine function itself. By isolating $sin(θ)$, we obtain $sin(θ) = \\frac{\\sqrt{2}}{2}$. This form of the equation is much easier to work with because it directly relates the sine of the angle to a specific value. Now, we can focus on finding the angles whose sine is equal to $\\frac{\\sqrt{2}}{2}$. This value is a well-known trigonometric ratio, which corresponds to specific angles on the unit circle. Isolating the trigonometric function is a crucial step in solving any trigonometric equation, as it sets the stage for finding the angles that satisfy the equation. This process is analogous to solving algebraic equations where isolating the variable is the first step towards finding the solution. With the sine function isolated, we can now proceed to identify the angles that have the desired sine value.

Identifying Reference Angles

Now that we have $sin(θ) = \\frac{\\sqrt{2}}{2}$, the next step is to identify the reference angles. Reference angles are acute angles formed by the terminal side of the angle and the x-axis. They help us determine the solutions in different quadrants. In this case, we know that $sin(\frac{\\pi}{4}) = \\frac{\\sqrt{2}}{2}$. Therefore, $\\frac{\\pi}{4}$ is our reference angle. The reference angle is the angle in the first quadrant that has the same sine value as the solutions we are looking for. Understanding reference angles is crucial because they allow us to find solutions in all four quadrants. The sine function is positive in the first and second quadrants, so we need to find angles in these quadrants that have a reference angle of $\\frac{\\pi}{4}$. In the first quadrant, the angle is simply the reference angle itself. In the second quadrant, we need to subtract the reference angle from $\\pi$ to find the solution. By using reference angles, we can systematically find all the solutions within the given interval. This approach simplifies the process of solving trigonometric equations and ensures that we don't miss any solutions.

Finding Solutions in the Correct Quadrants

Since $sin(θ) = \\frac{\\sqrt{2}}{2}$ is positive, we know that the solutions lie in the first and second quadrants, where the sine function is positive. We've already identified the reference angle as $\\frac{\\pi}{4}$. In the first quadrant, the solution is simply the reference angle, which is $θ = \\frac{\\pi}{4}$. To find the solution in the second quadrant, we subtract the reference angle from $\\pi$: $θ = \\pi - \\frac{\\pi}{4} = \\frac{3\\pi}{4}$. Therefore, the solutions in the interval $0 \\\leq \\\theta < 2\\\pi$ are $θ = \\frac{\\pi}{4}$ and $θ = \\frac{3\\pi}{4}$. It's important to verify that these solutions fall within the specified interval. Both $\\frac{\\pi}{4}$ and $\\frac{3\\pi}{4}$ are between 0 and $2\\\[Pi]$, so they are valid solutions. By considering the quadrants where the sine function is positive and using the reference angle, we have successfully found all the solutions to the equation within the given interval. This process highlights the importance of understanding the properties of trigonometric functions and their behavior in different quadrants.

Verifying the Solutions

After finding the potential solutions, it's crucial to verify them to ensure they satisfy the original equation. This step helps to catch any errors made during the solving process and confirms the accuracy of the answers. We have found two potential solutions: $θ = \\frac{\\pi}{4}$ and $θ = \\frac{3\\pi}{4}$. To verify these solutions, we substitute them back into the original equation, $-2sin(θ) = -\\sqrt{2}$. For $θ = \\frac{\\pi}{4}$, we have $-2sin(\\frac{\\pi}{4}) = -2(\\frac{\\sqrt{2}}{2}) = -\\sqrt{2}$, which satisfies the equation. For $θ = \\frac{3\\pi}{4}$, we have $-2sin(\\frac{3\\pi}{4}) = -2(\\frac{\\sqrt{2}}{2}) = -\\sqrt{2}$, which also satisfies the equation. Since both solutions satisfy the original equation, we can confidently conclude that they are the correct solutions. Verifying solutions is a best practice in mathematics, as it provides a final check on the accuracy of the work. This step is particularly important in trigonometry, where errors can easily occur due to the periodic nature of the functions and the multiple solutions that may exist.

Final Answer

In conclusion, the exact solutions to the equation $-2sin(θ) = -\\sqrt{2}$ on the interval $0 \\\leq \\\theta < 2\\\pi$ are $θ = \\frac{\\pi}{4}$ and $θ = \\frac{3\\pi}{4}$. These solutions were found by isolating the sine function, identifying the reference angle, determining the correct quadrants, and verifying the solutions. This process demonstrates a systematic approach to solving trigonometric equations, which can be applied to a wide range of problems. Understanding the properties of trigonometric functions and their behavior in different quadrants is essential for finding accurate solutions. By mastering these techniques, you can confidently solve trigonometric equations and apply them to various real-world applications. The ability to find exact solutions is a valuable skill in mathematics, science, and engineering, enabling precise calculations and a deeper understanding of trigonometric relationships. This comprehensive guide provides a clear and concise method for solving trigonometric equations, ensuring that you can tackle similar problems with ease and accuracy.

Answer:

$\theta = \frac{\pi}{4}, \frac{3\pi}{4}$