Solving Cos(2sinθ - 1) = 0 For 0° ≤ Θ ≤ 180°

This article delves into the intricate process of solving the trigonometric equation cos(2sinθ - 1) = 0 within the specified domain of 0° ≤ θ ≤ 180°. This problem combines trigonometric functions and algebraic manipulation, making it a valuable exercise for students and enthusiasts alike. We will break down the solution step-by-step, providing a clear and comprehensive explanation of the underlying principles and techniques involved. Understanding trigonometric equations is crucial in various fields, including physics, engineering, and computer graphics. Therefore, a solid grasp of these concepts is essential for anyone pursuing these disciplines. This detailed exploration aims to equip you with the necessary tools to tackle similar trigonometric problems with confidence and precision. Before diving into the specifics, let's establish a foundational understanding of trigonometric functions and their properties, especially the cosine function, which plays a central role in this equation.

Understanding the Cosine Function and Its Properties

The cosine function, denoted as cos(x), is a fundamental trigonometric function that relates an angle of a right triangle to the ratio of the adjacent side to the hypotenuse. Its values oscillate between -1 and 1, forming a periodic wave. The cosine function has several key properties that are essential for solving trigonometric equations. One crucial property is its periodicity, meaning it repeats its values after a certain interval. Specifically, cos(x) = cos(x + 360°), reflecting its cyclical nature. Another important characteristic is its symmetry: cos(x) = cos(-x), indicating that the cosine function is even. This symmetry helps simplify equations and find multiple solutions. To effectively solve the given equation, we must recall when the cosine function equals zero. Cos(x) = 0 when x is an odd multiple of 90°, such as 90°, 270°, 450°, and so on. In radians, this translates to x = (2n + 1)π/2, where n is an integer. This understanding forms the basis for our initial step in solving cos(2sinθ - 1) = 0. By equating the argument of the cosine function to the values where cosine is zero, we can transform the trigonometric equation into a simpler algebraic form. This foundation sets the stage for subsequent steps, where we will isolate sinθ and determine the possible values of θ within the given range. The interplay between the cosine function's properties and the algebraic manipulations will ultimately lead us to the solution.

Solving the Equation cos(2sinθ - 1) = 0

To solve the trigonometric equation cos(2sinθ - 1) = 0, our initial step involves recognizing that the cosine function equals zero at angles that are odd multiples of π/2 radians or 90 degrees. Therefore, we set the argument of the cosine function, (2sinθ - 1), equal to these values. This leads us to the equation 2sinθ - 1 = (2n + 1)π/2, where n is an integer. However, since we are working with degrees and aiming for solutions within the range of 0° ≤ θ ≤ 180°, it’s more practical to express the solutions in degrees. Thus, we have 2sinθ - 1 = 90° + 180°n, where n is an integer. Now, we need to isolate sinθ. By adding 1 to both sides of the equation, we get 2sinθ = 90° + 180°n + 1. Dividing both sides by 2 gives us sinθ = (90° + 180°n + 1) / 2. This expression for sinθ highlights the importance of considering different values of n to find solutions within our specified range. As n varies, sinθ will take on different values, and we must identify which values yield solutions for θ that fall between 0° and 180°. The next crucial step is to analyze the possible values of n and their corresponding sinθ values. We will focus on integer values of n that result in sinθ values between -1 and 1, as these are the valid range for the sine function. This process will help us narrow down the potential solutions for θ and ultimately determine the angles that satisfy the original equation. Remember, the sine function's behavior in different quadrants will play a significant role in this analysis.

Analyzing Possible Values of n and sinθ

Continuing from the equation sinθ = (90° + 180°n + 1) / 2, we must analyze different integer values of n to determine valid solutions for sinθ within the range of -1 to 1. Let's begin by testing a few values of n. For n = 0, we have sinθ = (90 + 1)/2 = 91/2, which is not valid since sinθ must be between -1 and 1. For n = -1, we have sinθ = (90 - 180 + 1)/2 = -89/2, also an invalid value. Now, let's consider a different approach. The cosine function is zero when its argument is of the form (2n + 1)π/2, where n is an integer. Therefore, 2sinθ - 1 = (2n + 1)π/2. Since we're working in degrees, this translates to 2sinθ - 1 = 90(2n + 1) degrees. Adding 1 to both sides gives 2sinθ = 90(2n + 1) + 1, and dividing by 2 yields sinθ = [90(2n + 1) + 1] / 2. We need to find integer values of n for which -1 ≤ sinθ ≤ 1. Substituting different values of n, we find that for n = -1, sinθ = [90(-1) + 1]/2 = -89/2, which is less than -1. For n = 0, 2sinθ - 1 = 90°, which gives 2sinθ = 91°. This seems incorrect as the argument of the cosine function should be in radians, not degrees, when setting it equal to the values where cosine is zero. Let's go back to our original premise and correct the approach. We know that cos(2sinθ - 1) = 0 implies that 2sinθ - 1 = (2n + 1)π/2, where n is an integer. Since the argument of the cosine function is in radians, we'll stick to radians for now. This means 2sinθ - 1 = (2n + 1)π/2. Adding 1 to both sides gives 2sinθ = 1 + (2n + 1)π/2, and dividing by 2, we get sinθ = [1 + (2n + 1)π/2] / 2. Now, we need to find integer values of n such that -1 ≤ sinθ ≤ 1. This revised approach will help us accurately determine the values of θ within the given range.

Finding Valid Solutions for θ

To find the valid solutions for θ, we continue with our equation sinθ = [1 + (2n + 1)π/2] / 2 and seek integer values of n that result in sinθ values between -1 and 1. This range is crucial because the sine function's output always falls within these bounds. Let's analyze the equation further. Multiplying both the numerator and denominator by 2, we get sinθ = (2 + (2n + 1)π) / 4. Now, we need to find the values of n for which -1 ≤ (2 + (2n + 1)π) / 4 ≤ 1. Multiplying all parts of the inequality by 4 gives us -4 ≤ 2 + (2n + 1)π ≤ 4. Subtracting 2 from all parts, we have -6 ≤ (2n + 1)π ≤ 2. Dividing by π, we get -6/π ≤ 2n + 1 ≤ 2/π. Approximating π as 3.14159, we have -6/3.14159 ≈ -1.91 and 2/3.14159 ≈ 0.637. Thus, our inequality becomes -1.91 ≤ 2n + 1 ≤ 0.637. Subtracting 1 from all parts, we get -2.91 ≤ 2n ≤ -0.363. Finally, dividing by 2, we have -1.455 ≤ n ≤ -0.1815. Since n must be an integer, the only possible value for n is -1. Now, substitute n = -1 back into the equation for sinθ: sinθ = [2 + (2(-1) + 1)π] / 4 = (2 - π) / 4. Approximating π as 3.14159, we have sinθ ≈ (2 - 3.14159) / 4 ≈ -1.14159 / 4 ≈ -0.2854. Now that we have a value for sinθ, we can find the corresponding values of θ within the range 0° ≤ θ ≤ 180°. Since sinθ is negative, we know that θ must lie in the third or fourth quadrant. However, our range is limited to 0° to 180°, which includes the first and second quadrants. To find the principal value, we take the inverse sine: θ = arcsin(-0.2854) ≈ -16.57°. This angle is not within our desired range. To find the angles within the range 0° ≤ θ ≤ 180°, we use the properties of the sine function. In the second quadrant, sin(180° - θ) = sinθ. However, since our principal value is negative, we need to find an angle in the third or fourth quadrant first, and then adjust. Since we are only considering 0° ≤ θ ≤ 180°, and sine is negative, there are no solutions in the given range. However, let's re-examine our initial steps to ensure accuracy. We had sinθ = [1 + (2n + 1)π/2] / 2. When 2sinθ-1=π/2, we get sinθ=1.285, which is not between -1 and 1. Therefore, there's an error in our calculation. We'll correct it in the next section by revisiting our initial approach and identifying the correct method for finding the solutions.

Correcting the Approach and Finding the Solution

Let's revisit our initial approach to solving cos(2sinθ - 1) = 0 for 0° ≤ θ ≤ 180°. The critical point is understanding that cos(x) = 0 when x = (2n + 1)π/2, where n is an integer. Therefore, we must have 2sinθ - 1 = (2n + 1)π/2. Now, we need to express this in degrees for easier calculation within the given range. We know that π radians is 180 degrees, so π/2 radians is 90 degrees. Thus, we have 2sinθ - 1 = (2n + 1)90°. Adding 1 to both sides gives 2sinθ = 1 + (2n + 1)90°, and dividing by 2, we get sinθ = [1 + (2n + 1)90°] / 2. Now we analyze different integer values of n to find sinθ values between -1 and 1. For n = 0, sinθ = (1 + 90) / 2 = 91/2, which is not valid. For n = -1, sinθ = (1 - 90) / 2 = -89/2, which is also not valid. However, we made an error in converting radians to degrees earlier. We should stick to radians for the argument of the cosine function. Let's go back to 2sinθ - 1 = (2n + 1)π/2. We need to find integer values of n that make this true. Let's try n = 0: 2sinθ - 1 = π/2. Adding 1 to both sides, we get 2sinθ = 1 + π/2. Dividing by 2, we have sinθ = (1 + π/2) / 2. Approximating π as 3.14159, we have sinθ ≈ (1 + 1.5708) / 2 ≈ 2.5708 / 2 ≈ 1.2854. This value is greater than 1, so it's not valid. Let's try n = -1: 2sinθ - 1 = -π/2. Adding 1 to both sides, we get 2sinθ = 1 - π/2. Dividing by 2, we have sinθ = (1 - π/2) / 2. Approximating π as 3.14159, we have sinθ ≈ (1 - 1.5708) / 2 ≈ -0.5708 / 2 ≈ -0.2854. This value is between -1 and 1, so it's a possible solution. Now we find θ: θ = arcsin(-0.2854). The principal value is approximately -16.57°. Since we want solutions in the range 0° ≤ θ ≤ 180°, we need to find the angles in the second quadrant. However, the sine is negative, so there are no solutions in the first or second quadrant. Let's check n = -2: 2sinθ - 1 = -3π/2. Adding 1, 2sinθ = 1 - 3π/2. Dividing by 2, sinθ = (2 - 3π) / 4. Approximating, sinθ ≈ (2 - 9.42478) / 4 ≈ -7.42478 / 4 ≈ -1.856. This is outside the range [-1, 1], so it's not valid. Therefore, the only valid case we found is when n = -1, giving sinθ ≈ -0.2854. However, since the sine function is negative, there are no solutions for θ in the range 0° ≤ θ ≤ 180°. Therefore, the final answer is that there are no solutions to the equation within the specified range.

Conclusion: No Solution within the Given Range

In conclusion, after a detailed step-by-step analysis, we determined that the trigonometric equation cos(2sinθ - 1) = 0 has no solutions within the range of 0° ≤ θ ≤ 180°. This determination involved a careful examination of the cosine and sine functions, their properties, and the interplay between algebraic manipulations and trigonometric principles. We started by recognizing that the cosine function equals zero at odd multiples of π/2 radians and setting the argument of the cosine function accordingly. We then explored different values of n to find potential solutions for sinθ within the valid range of -1 to 1. Throughout the process, we corrected errors in our approach, particularly concerning the conversion between radians and degrees, ensuring the accuracy of our calculations. Ultimately, we found that the only possible value for sinθ, approximately -0.2854, corresponded to an angle outside the specified range of 0° ≤ θ ≤ 180°. This negative sine value implies angles in the third and fourth quadrants, which lie beyond our considered interval. This problem underscores the importance of a thorough and methodical approach to solving trigonometric equations. It highlights the necessity of understanding trigonometric functions' properties, careful algebraic manipulation, and attention to detail in calculations. While the equation had no solutions in this particular range, the process of solving it has provided valuable insights into the behavior of trigonometric functions and the techniques required to tackle similar problems effectively. The absence of a solution, in this case, is as informative as finding one, as it reinforces the constraints and characteristics of trigonometric functions.