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

In this comprehensive guide, we will delve into the process of finding the exact solutions for the trigonometric equation -2sin(θ) = -√2 within the interval 0 ≤ θ < 2π. Trigonometric equations play a crucial role in various fields, including physics, engineering, and mathematics. Mastering the techniques to solve these equations is essential for anyone pursuing these disciplines. This article will provide a step-by-step approach, ensuring a clear understanding of the concepts involved. We'll explore the fundamental principles of trigonometry, focusing on the sine function and its behavior within the unit circle. By the end of this guide, you'll be well-equipped to tackle similar problems and confidently determine the exact solutions of trigonometric equations.

This exploration into trigonometric equations begins with a foundational understanding of trigonometric functions, particularly the sine function. The sine function, denoted as sin(θ), relates an angle θ to the ratio of the opposite side to the hypotenuse in a right-angled triangle. In the context of the unit circle, where the radius is 1, sin(θ) represents the y-coordinate of the point where the terminal side of the angle θ intersects the circle. The unit circle provides a visual representation of trigonometric functions, making it easier to understand their periodic nature and values at different angles. Understanding the sine function's behavior across the unit circle is critical for solving trigonometric equations. The sine function oscillates between -1 and 1, completing a full cycle over an interval of 2π radians (or 360 degrees). This periodic nature means that the sine function repeats its values at regular intervals, leading to multiple solutions for trigonometric equations within a given range. To effectively solve equations involving sine, it is important to know the angles at which sin(θ) takes on specific values, such as 0, 1, -1, √2/2, √3/2, and 1/2. These values are frequently encountered in trigonometric problems, and familiarity with them will greatly simplify the solution process. Additionally, understanding the symmetry properties of the sine function, such as sin(θ) = sin(π - θ), is crucial for finding all possible solutions within a given interval.

The first crucial step in solving the given trigonometric equation, -2sin(θ) = -√2, is to isolate the sine function. This involves algebraic manipulation to get sin(θ) by itself on one side of the equation. By isolating the trigonometric function, we simplify the equation and make it easier to determine the angles that satisfy the condition. This isolation process is fundamental to solving any trigonometric equation, as it allows us to directly relate the trigonometric function to a specific value. To isolate sin(θ) in the equation -2sin(θ) = -√2, we divide both sides of the equation by -2. This operation effectively cancels out the coefficient of sin(θ), leaving the sine function alone on the left-hand side. The resulting equation, sin(θ) = √2/2, is much simpler to work with. It directly tells us that we are looking for angles θ whose sine value is √2/2. This step is a critical prerequisite for finding the solutions, as it transforms the original equation into a standard form that can be easily solved using known trigonometric values and properties. Once the sine function is isolated, we can proceed to identify the angles within the specified interval that satisfy the equation. The equation sin(θ) = √2/2 is a classic trigonometric problem with well-known solutions, which makes the subsequent steps more straightforward. Isolating the trigonometric function not only simplifies the equation but also allows us to apply our knowledge of trigonometric values and the unit circle to find the solutions efficiently.

After isolating the sine function and obtaining the equation sin(θ) = √2/2, the next step is to determine the reference angle. The reference angle is the acute angle formed between the terminal side of the angle θ and the x-axis. It provides a fundamental building block for finding all possible solutions within the given interval. The reference angle is always a positive acute angle, typically between 0 and π/2 radians (or 0 and 90 degrees). It allows us to relate the trigonometric value of any angle to its equivalent value in the first quadrant, where all trigonometric functions are positive. For the equation sin(θ) = √2/2, we need to identify the acute angle whose sine is √2/2. This is a standard trigonometric value that students are often expected to memorize. The angle whose sine is √2/2 is π/4 radians (or 45 degrees). This angle serves as our reference angle. The reference angle, π/4, is crucial because it allows us to find all angles θ that have the same sine value, considering the periodic and symmetric properties of the sine function. In other words, we know that sin(π/4) = √2/2, but there might be other angles in different quadrants that also have a sine of √2/2. These other angles will be related to the reference angle, and determining them is the next step in solving the trigonometric equation. The reference angle simplifies the process of finding all solutions by providing a known angle that we can use as a basis for identifying other angles with the same sine value. It is an essential concept in solving trigonometric equations and understanding the behavior of trigonometric functions.

Having determined the reference angle, the next critical step is to identify the quadrants in which the sine function is positive. This is crucial because the sine function's sign varies across the four quadrants of the coordinate plane. Understanding where sine is positive helps us pinpoint the angles within the interval 0 ≤ θ < 2π that satisfy our equation, sin(θ) = √2/2. The sine function corresponds to the y-coordinate on the unit circle. Therefore, sine is positive in the quadrants where the y-coordinate is positive. These are the first and second quadrants. In the first quadrant, both the x and y coordinates are positive, so all trigonometric functions (sine, cosine, and tangent) are positive. This quadrant is the simplest to work with as the angles directly correspond to their trigonometric values. In the second quadrant, the y-coordinate is positive while the x-coordinate is negative. This means that sine is positive, cosine is negative, and tangent is negative. Knowing that sine is positive in the first and second quadrants narrows down our search for solutions to these two regions of the unit circle. We already have the reference angle, π/4, which lies in the first quadrant. Now, we need to find the corresponding angle in the second quadrant that also has a sine value of √2/2. By understanding the quadrants where sine is positive, we can efficiently identify all possible solutions to the trigonometric equation within the specified interval. This step is a key component in the process of solving trigonometric equations, as it allows us to use the symmetry and periodicity of trigonometric functions to our advantage.

With the reference angle (π/4) and the quadrants where sine is positive (first and second) identified, we can now find the solutions for θ in the interval 0 ≤ θ < 2π. This involves determining the angles in each of these quadrants that have a sine value of √2/2. In the first quadrant, the angle is simply the reference angle itself. So, one solution is θ = π/4. This is because the reference angle is defined as the acute angle formed between the terminal side of θ and the x-axis, and in the first quadrant, this angle directly corresponds to θ. To find the solution in the second quadrant, we need to use the property that sin(θ) = sin(π - θ). This property arises from the symmetry of the sine function about the y-axis. In the second quadrant, the angle θ is given by π minus the reference angle. Therefore, the second solution is θ = π - π/4. Calculating this gives us θ = 3π/4. So, the two angles in the interval 0 ≤ θ < 2π that satisfy the equation sin(θ) = √2/2 are π/4 and 3π/4. These are the exact solutions we are looking for. To summarize, we used the reference angle to find the solution in the first quadrant, and the symmetry property of the sine function to find the solution in the second quadrant. This approach leverages the fundamental properties of trigonometric functions and the unit circle to efficiently determine the solutions within the specified interval. Finding solutions in the specified interval is the culmination of the previous steps, and it provides the final answer to the trigonometric equation. These solutions represent the angles that, when plugged into the original equation, will make the equation true.

Having found the solutions θ = π/4 and θ = 3π/4 within the interval 0 ≤ θ < 2π, the final step is to express these solutions clearly and accurately. Presenting the solutions in the correct format is essential for conveying the answer effectively. In this case, the solutions are expressed in radians, which is the standard unit for measuring angles in mathematical contexts. The solutions π/4 and 3π/4 represent the angles in radians that satisfy the given trigonometric equation, -2sin(θ) = -√2. These values correspond to angles in the first and second quadrants, respectively, where the sine function is positive. To ensure clarity, it is common practice to list the solutions in ascending order within the specified interval. In this case, the solutions are already in ascending order: π/4 and 3π/4. It is also important to double-check that these solutions fall within the given interval, 0 ≤ θ < 2π. Both π/4 and 3π/4 are within this range, so they are valid solutions. Expressing the solutions correctly involves not only finding the correct values but also presenting them in a manner that is easy to understand and interpret. In many cases, solutions may be required in different units (e.g., degrees), or as a general solution that includes all possible angles, considering the periodic nature of trigonometric functions. However, for this specific problem, the solutions π/4 and 3π/4 in radians are the final and complete answer. These solutions can be used in any application or problem where this trigonometric equation arises, providing a precise and accurate result.

In conclusion, we have successfully navigated the process of finding the exact solutions for the trigonometric equation -2sin(θ) = -√2 within the interval 0 ≤ θ < 2π. This journey involved a series of logical steps, each building upon the previous one, to arrive at the final answer. We began by isolating the sine function, transforming the original equation into the simpler form sin(θ) = √2/2. This critical step set the stage for the rest of the solution process. Next, we determined the reference angle, π/4, which served as the foundation for identifying all possible solutions. The reference angle allowed us to relate the trigonometric value to a known angle in the first quadrant, simplifying the search for other solutions. We then identified the quadrants where the sine function is positive, which are the first and second quadrants. This step narrowed down the possible locations of solutions within the unit circle, making the problem more manageable. Using the reference angle and the knowledge of positive sine quadrants, we found the solutions within the specified interval: θ = π/4 and θ = 3π/4. These solutions represent the angles that satisfy the given equation and fall within the range of 0 to 2π. Finally, we expressed the solutions clearly and accurately, ensuring that the answer is presented in a manner that is easy to understand and interpret. The solutions, π/4 and 3π/4, represent the exact angles that make the original equation true. This process demonstrates the systematic approach required to solve trigonometric equations, emphasizing the importance of understanding trigonometric functions, their properties, and the unit circle. By mastering these techniques, one can confidently tackle a wide range of trigonometric problems and apply these skills in various mathematical and scientific contexts.