Solving Trigonometric Equations: A Step-by-Step Guide

Hey everyone! Today, we're diving into the world of trigonometry and tackling an interesting equation. We'll be solving for $ heta$ in the equation $tan heta = \sqrt{2} \sin heta$ within the interval $0 \leq \theta < 2\pi$. Don't worry, it might look a little intimidating at first, but we'll break it down step by step to make sure you understand everything. This guide will walk you through the process, providing clear explanations, and helping you build a solid understanding of how to solve these kinds of trigonometric problems. Let's get started and unravel this equation together! Ready to jump in and solve this together? Let's go!

Understanding the Problem: The Equation and the Interval

Alright, before we get our hands dirty with calculations, let's make sure we're on the same page. The equation we're dealing with is $tan heta = \sqrt{2} \sin heta$. Remember, $tan heta$ is the same as $\frac{\sin \theta}{\cos \theta}$. Our goal is to find all the values of $ heta$ that make this equation true. But there's a catch – we're only looking for solutions within a specific range, or interval: $0 \leq \theta < 2\pi$. This means $ heta$ can be anything from 0 to, but not including, $2\pi$ radians (which is a full circle). This is important because trigonometric functions are periodic, meaning they repeat their values over and over. By restricting the interval, we are limiting ourselves to a single cycle of the function, which makes finding the solutions much more manageable. The restriction ensures we only get unique solutions within that specific range. Keep this in mind, as it helps us determine which solutions are valid and which ones we need to discard. We will explore each step carefully, ensuring you understand not just how to solve the equation, but also why we approach it in this manner. Ready? Let's decode this!

Step 1: Rewriting the Equation

Okay, guys, the first thing we're gonna do is rewrite the equation. It's often helpful to get everything in terms of sine and cosine, since those are the fundamental trigonometric functions. Let's start with $tan heta = \sqrt{2} \sin heta$. As we mentioned before, we know that $tan \theta = \frac{\sin \theta}{\cos \theta}$. So, let's substitute that into the equation: $\frac{\sin \theta}{\cos \theta} = \sqrt{2} \sin \theta$. See, it’s not so bad, right? We've already simplified the equation a bit just by doing this one substitution. This step is about transforming the equation into a form that's easier to manipulate. It is very important as it gives us a clear path to follow. This conversion paves the way for the next phase. The aim here is to make the equation less complex, by using the most common form of trigonometric functions, which are sine and cosine. This will now help us in solving the equation with ease.

Isolating terms and factoring

Now, let's move everything to one side of the equation. We'll subtract $\sqrt{2} \sin \theta$ from both sides to get: $\frac{\sin \theta}{\cos \theta} - \sqrt{2} \sin \theta = 0$. Now, we want to combine these terms. To do this, we'll factor out a $\sin \theta$: $\sin \theta (\frac{1}{\cos \theta} - \sqrt{2}) = 0$. This is a super important step because it transforms the equation into a form where we can easily identify potential solutions. When we have a product of factors equal to zero, we know that at least one of the factors must be zero. This is a crucial principle that simplifies the process of finding solutions. That means, to solve this, we can set each factor equal to zero and solve for $ heta$. This strategy significantly simplifies the problem, making it easier to solve.

Step 2: Solving for $\theta$

Now that we've got our equation in a factored form, it's time to find the solutions. Remember, we have two factors: $\sin \theta$ and $\frac{1}{\cos \theta} - \sqrt{2}$. We'll set each factor to zero and solve for $\theta$. Let's start with $\sin \theta = 0$. We need to find the values of $\theta$ in the interval $0 \leq \theta < 2\pi$ where the sine function equals zero. If you recall the unit circle, $\sin \theta$ represents the y-coordinate. So, $\sin \theta = 0$ when the y-coordinate is zero, which happens at $\theta = 0$ and $\theta = \pi$. There is one part down, not so bad right?

Solving the second factor

Next, let’s solve the second part. Now we are going to set $\frac{1}{\cos \theta} - \sqrt{2} = 0$. Let’s rearrange and solve for $\cos \theta$. First, add $\sqrt{2}$ to both sides: $\frac{1}{\cos \theta} = \sqrt{2}$. Now, take the reciprocal of both sides: $\cos \theta = \frac{1}{\sqrt{2}}$. This is the same as $\cos \theta = \frac{\sqrt{2}}{2}$. So, we're looking for angles where the cosine function equals $\frac{\sqrt{2}}{2}$. Recall that the cosine function represents the x-coordinate on the unit circle. This occurs at $\theta = \frac{\pi}{4}$ and $\theta = \frac{7\pi}{4}$. Therefore, for this part of the solution, we've found two more solutions! With these two, we have all possible values of $\theta$. Finding the solutions often involves recalling the unit circle or using a calculator to find the angles that satisfy the equation. This part can be tricky, so make sure to take your time and double-check your work!

Step 3: Listing the Solutions

Alright, guys, let's put everything together. We've found the following solutions for $\theta$ in the interval $0 \leq \theta < 2\pi$: $\theta = 0$, $\theta = \pi$, $\theta = \frac{\pi}{4}$, and $\theta = \frac{7\pi}{4}$. Congratulations, we have successfully solved the equation! The four found values are our final answer. It's a great feeling to arrive at the conclusion after all the hard work. We can confidently say that these are all the values of $\theta$ that satisfy the original equation within the given interval. To recap, we started by simplifying the equation. We then isolated the variables, factored the equation, and solved for theta. Remember, practicing more problems will help to solidify your understanding. It's a journey, and with each problem, we become more confident and skilled. Keep up the great work and always remember to enjoy the process of learning.

Step 4: Verification and Conclusion

Checking our answers

To make sure we've done everything right, it's always a good idea to check our solutions. We'll plug each of our solutions back into the original equation $tan \theta = \sqrt{2} \sin \theta$. Let's start with $\theta = 0$: $tan(0) = 0$ and $\sqrt{2} \sin(0) = 0$. So, $0 = 0$. That checks out! Next, for $\theta = \pi$: $tan(\pi) = 0$ and $\sqrt{2} \sin(\pi) = 0$. Again, $0 = 0$. Another one works! For $\theta = \frac{\pi}{4}$: $tan(\frac{\pi}{4}) = 1$ and $\sqrt{2} \sin(\frac{\pi}{4}) = \sqrt{2} * \frac{\sqrt{2}}{2} = 1$. Awesome! Finally, for $\theta = \frac{7\pi}{4}$: $tan(\frac{7\pi}{4}) = -1$ and $\sqrt{2} \sin(\frac{7\pi}{4}) = \sqrt{2} * \frac{-\sqrt{2}}{2} = -1$. Everything checks out! This step is crucial because it ensures our solutions are valid. Sometimes, we might encounter extraneous solutions, which are solutions that arise during the solving process but don't satisfy the original equation. By checking our answers, we can identify and discard any extraneous solutions. So by now, we've checked all of our solutions and made sure they're correct. It's a great practice that ensures accuracy and reinforces our understanding of the concepts.

Final Answer

Therefore, the solutions to the equation $tan \theta = \sqrt{2} \sin \theta$ in the interval $0 \leq \theta < 2\pi$ are $\theta = 0$, $\theta = \pi$, $\theta = \frac{\pi}{4}$, and $\theta = \frac{7\pi}{4}$. We have successfully navigated through the equation. And that's a wrap, everyone! We have reached the final conclusion. We hope that this step-by-step guide has been helpful and that you now feel more confident in solving trigonometric equations. This problem showcases some important techniques for solving trig equations, like rewriting using sine and cosine, factoring, and carefully considering the given interval. Keep practicing, and you'll become a pro in no time! Remember, the key to mastering any math concept is practice. So, try solving similar equations on your own, and you'll solidify your understanding. See you in the next one! Keep learning and keep exploring the amazing world of mathematics! Bye!