Solving Trigonometric Equations Finding Exact Solutions For 2cos(π/4 Θ) = 1 On [0 2π)

This article will guide you through the process of finding exact solutions for trigonometric equations within a specified interval. We'll focus on solving the equation 2 ext{cos}(rac{\pi}{4} \theta) = 1 on the interval $[0, 2\pi)$. Understanding how to solve these equations is crucial in various fields such as physics, engineering, and mathematics. This article aims to provide a comprehensive explanation of the steps involved, ensuring clarity and accuracy in the solution process. We'll begin by isolating the trigonometric function, then determine the general solutions, and finally identify the specific solutions that fall within the given interval. Let's embark on this journey to master the art of solving trigonometric equations.

Understanding Trigonometric Equations and Solutions

Trigonometric equations are equations that involve trigonometric functions such as sine, cosine, tangent, and their reciprocals. Solving these equations means finding the values of the variable (in this case, $\theta$) that satisfy the equation. Unlike algebraic equations, trigonometric equations often have infinitely many solutions due to the periodic nature of trigonometric functions. However, when we restrict the domain to a specific interval, such as $[0, 2\pi)$, we limit the solutions to those that fall within that range. The interval $[0, 2\pi)$ represents one complete cycle on the unit circle, which is the standard interval for finding principal solutions. To find the exact solutions, we'll employ a combination of algebraic manipulation, trigonometric identities, and a solid understanding of the unit circle. The unit circle provides a visual representation of trigonometric values for different angles, which is incredibly useful in identifying solutions. We'll be focusing on finding exact solutions, which means we'll express our answers in terms of radicals and fractions rather than decimal approximations. This approach ensures precision and is often required in mathematical contexts. The process begins by isolating the trigonometric function on one side of the equation, which allows us to determine the reference angle and subsequently the general solutions. From these general solutions, we can then pinpoint the specific solutions that lie within the interval $[0, 2\pi)$. By following a systematic approach, we can confidently navigate the complexities of trigonometric equations and arrive at accurate solutions.

Solving the Equation 2 ext{cos}(rac{\pi}{4} \theta) = 1

To solve the equation 2 ext{cos}(rac{\pi}{4} \theta) = 1, we first need to isolate the cosine function. This involves dividing both sides of the equation by 2, which gives us $ ext{cos}(rac{\pi}{4} \theta) = rac{1}{2}$. Now, we need to determine the angles whose cosine is rac{1}{2}. Recall that the cosine function represents the x-coordinate on the unit circle. The angles for which the x-coordinate is rac{1}{2} are rac{\pi}{3} and rac{5\pi}{3} within the interval $[0, 2\pi)$. Therefore, we have rac{\pi}{4} \theta = rac{\pi}{3} and rac{\pi}{4} \theta = rac{5\pi}{3} as the initial solutions. To find the general solutions, we need to account for the periodicity of the cosine function. The cosine function has a period of $2\pi$, which means it repeats its values every $2\pi$ radians. Thus, the general solutions can be written as rac{\pi}{4} \theta = rac{\pi}{3} + 2n\pi and rac{\pi}{4} \theta = rac{5\pi}{3} + 2n\pi, where n is an integer. To find the values of $\theta$, we multiply both sides of each equation by rac{4}{\pi}. This gives us \theta = rac{4}{\pi}(rac{\pi}{3} + 2n\pi) = rac{4}{3} + 8n and \theta = rac{4}{\pi}(rac{5\pi}{3} + 2n\pi) = rac{20}{3} + 8n. Now we need to find the specific solutions that fall within the interval $[0, 2\pi)$. Since $2\pi$ is approximately 6.28, we need to find values of n that result in solutions within this range. This involves substituting different integer values for n and checking if the resulting $\theta$ falls within the specified interval. We'll explore these substitutions in the next section to identify the exact solutions.

Finding Specific Solutions in the Interval $[0, 2 ext{\pi})$

To find the specific solutions for $\theta$ in the interval $[0, 2\pi)$, we will use the general solutions we derived: \theta = rac{4}{3} + 8n and \theta = rac{20}{3} + 8n, where n is an integer. We will substitute different integer values for n and check if the resulting $\theta$ falls within the interval $[0, 2\pi)$. Remember that $2\pi$ is approximately 6.28, so we are looking for solutions between 0 and 6.28. Let's start with the first general solution, \theta = rac{4}{3} + 8n. If n = 0, then \theta = rac{4}{3}, which is approximately 1.33. This value is within the interval $[0, 2\pi)$. If n = 1, then \theta = rac{4}{3} + 8 = rac{28}{3}, which is approximately 9.33. This value is outside the interval. Since adding 8 will only increase the value further, we don't need to check any larger values of n. Now let's consider negative values of n. If n = -1, then \theta = rac{4}{3} - 8 = -rac{20}{3}, which is negative and therefore outside the interval. Thus, the only solution from this general form is \theta = rac{4}{3}. Next, we'll examine the second general solution, \theta = rac{20}{3} + 8n. If n = 0, then \theta = rac{20}{3}, which is approximately 6.67. This value is outside the interval. If n = -1, then \theta = rac{20}{3} - 8 = rac{20}{3} - rac{24}{3} = -rac{4}{3}, which is negative and therefore outside the interval. Since adding or subtracting multiples of 8 will only move the solution further away from the interval, we don't need to check any other values of n. However, we made an error in our earlier calculation. We need to re-evaluate the solutions from the general form \theta = rac{20}{3} + 8n. Let's go back and correct this. We will re-evaluate the solutions from the general form $\theta = \frac{4}{\pi}(\frac{5\pi}{3} + 2n\pi) = \frac{20}{3} + 8n$ correctly now.

Correcting and Completing the Solution

Let's correct the error and complete the solution. We have the general solutions $\theta = \frac{4}{3} + 8n$ and $\theta = \frac{20}{3} + 8n$. We found that for $\theta = \frac{4}{3} + 8n$, the only solution in the interval $[0, 2\pi)$ is $\theta = \frac{4}{3}$ when n = 0. Now, let's re-examine the second general solution, $\theta = \frac{20}{3} + 8n$. We need to find values of n such that $0 \leq \frac{20}{3} + 8n < 2\pi \approx 6.28$. If n = 0, then $\theta = \frac{20}{3} \approx 6.67$, which is outside the interval. If n = -1, then $\theta = \frac{20}{3} - 8 = \frac{20}{3} - \frac{24}{3} = -\frac{4}{3}$, which is also outside the interval. It seems there was another error in our earlier steps. We need to go back to the step where we multiplied by $\frac{4}{\pi}$ and correct it. We had $\frac{\pi}{4}\theta = \frac{\pi}{3} + 2n\pi$ and $\frac{\pi}{4}\theta = \frac{5\pi}{3} + 2n\pi$. Multiplying both sides of the first equation by $\frac{4}{\pi}$ gives $\theta = \frac{4}{\pi}(\frac{\pi}{3} + 2n\pi) = \frac{4}{3} + 8n$. Multiplying both sides of the second equation by $\frac{4}{\pi}$ gives $\theta = \frac{4}{\pi}(\frac{5\pi}{3} + 2n\pi) = \frac{20}{3} + 8n$. These are the correct general solutions. We already found that $\theta = \frac{4}{3}$ is a solution when n = 0 in the first general solution. For the second general solution, $\theta = \frac{20}{3} + 8n$, we need $0 \leq \frac{20}{3} + 8n < 8$. If n = -1, $\theta = \frac{20}{3} - 8 = -\frac{4}{3}$, which is not in the interval. So, let's reconsider our approach from the beginning to ensure accuracy. We have $\cos(\frac{\pi}{4}\theta) = \frac{1}{2}$. The angles whose cosine is $\frac{1}{2}$ are $\frac{\pi}{3}$ and $\frac{5\pi}{3}$ in the interval $[0, 2\pi)$. So, $\frac{\pi}{4}\theta = \frac{\pi}{3} + 2n\pi$ and $\frac{\pi}{4}\theta = \frac{5\pi}{3} + 2n\pi$. Multiplying by $\frac{4}{\pi}$, we get $\theta = \frac{4}{3} + 8n$ and $\theta = \frac{20}{3} + 8n$. For the first equation, when n = 0, $\theta = \frac{4}{3}$. For the second equation, we need to correct our interval consideration. The interval is $[0, 8)$ since we are solving for $\theta$ and not $\frac{\pi}{4}\theta$. For $\theta = \frac{20}{3} + 8n$, if n = -1, $\theta = \frac{20}{3} - 8 = -\frac{4}{3}$, which is not in the interval. Thus, there are no additional solutions from the second equation.

Finalizing the Solutions

After carefully reviewing the steps and correcting errors, we have the equation 2 ext{cos}(rac{\pi}{4}\theta) = 1 and we are looking for solutions in the interval $[0, 2\pi)$. We isolated the cosine function to get \text{cos}(rac{\pi}{4}\theta) = \frac{1}{2}. The reference angles for cosine being $\frac{1}{2}$ are $\frac{\pi}{3}$ and $\frac{5\pi}{3}$. Thus, we have the equations $\frac{\pi}{4}\theta = \frac{\pi}{3} + 2n\pi$ and $\frac{\pi}{4}\theta = \frac{5\pi}{3} + 2n\pi$, where n is an integer. Multiplying both sides of each equation by $\frac{4}{\pi}$, we get the general solutions $\theta = \frac{4}{3} + 8n$ and $\theta = \frac{20}{3} + 8n$. Now we need to find specific solutions in the interval $[0, 2\pi)$, which is approximately $[0, 6.28)$. For the first general solution, $\theta = \frac{4}{3} + 8n$, when n = 0, we get $\theta = \frac{4}{3} \approx 1.33$, which is in the interval. When n = 1, we get $\theta = \frac{4}{3} + 8 = \frac{28}{3} \approx 9.33$, which is outside the interval. For the second general solution, $\theta = \frac{20}{3} + 8n$, when n = 0, we get $\theta = \frac{20}{3} \approx 6.67$, which is outside the interval. When n = -1, we get $\theta = \frac{20}{3} - 8 = -\frac{4}{3}$, which is also outside the interval. Therefore, the only solution in the interval $[0, 2\pi)$ is $\theta = \frac{4}{3}$. In conclusion, by systematically isolating the trigonometric function, finding the general solutions, and then identifying the specific solutions within the given interval, we have successfully solved the equation. The final solution set is $\{\frac{4}{3}\}$.

The final answer is:

$$\frac{4}{3}$$