Unlocking Solutions: Solving |sin X| = |sin 2x|

Hey math enthusiasts! Today, we're diving deep into the intriguing world of trigonometry to solve the absolute value equation: |sin x| = |sin 2x|. This equation might seem a bit daunting at first glance, with those absolute values throwing a wrench in the works. But fear not, guys! We'll break it down step-by-step, using a combination of trigonometric identities, case analysis, and a dash of mathematical intuition to conquer this problem. This isn't just about finding the answer; it's about understanding the why behind each step, making the journey as rewarding as the destination. So, buckle up, grab your pens and paper, and let's embark on this exciting mathematical adventure together. Our goal is to find all the values of x that satisfy this equation, which means we need to consider all possible scenarios where the absolute values might affect the sign of the sine function. Let's make sure that we have a solid understanding of the basics before we start working on the problem. Remember, the absolute value of a number is its distance from zero, so it's always non-negative. This means we'll be dealing with both positive and negative values of the sine function, and that's where the fun (and the cases) begin!

Demystifying Absolute Values and Trigonometric Identities

Before we jump into the core of the solution, let's refresh our memories on a few key concepts. Firstly, what exactly does the absolute value symbol, represented by | |, do? As we mentioned earlier, it essentially turns any negative value into its positive counterpart, leaving positive values unchanged. For example, |-3| = 3 and |5| = 5. Secondly, we'll need to remember some essential trigonometric identities. In this case, the double-angle formula for sine, which states that sin 2x = 2 sin x cos x, will be our most valuable weapon. This identity allows us to rewrite the equation in a way that simplifies our analysis. Understanding this identity is crucial because it helps us to deal with the sin 2x term directly. It transforms the equation from something that looks complex into a form that's easier to manipulate and solve. With these tools in our arsenal, we're ready to start dissecting the original equation. We'll methodically examine different scenarios and use our knowledge to systematically derive the solutions. The key here is not to rush but to analyze each case meticulously. This structured approach ensures that we don't miss any possible solutions and that our reasoning is sound. Remember, solving mathematical problems is like detective work, where each step reveals a clue that leads us closer to the truth. Let's get started!

Case Analysis: Breaking Down the Equation

Now, let's get into the heart of the matter: solving the equation |sin x| = |sin 2x|. The presence of absolute values suggests that we will need to consider different cases depending on the signs of sin x and sin 2x. This is where our case analysis comes into play. Think of it as breaking down a complex problem into more manageable chunks. We will explore each possible combination of signs and solve for x in each case. This structured approach helps us to methodically consider all possibilities and ensure that we don't miss any solutions. Case analysis is a powerful technique in mathematics, and it allows us to tackle even the most complicated equations step by step. We have to be careful with the details of each step to guarantee the correctness of our solution. Let's get started, shall we?

Case 1: sin x and sin 2x are both non-negative

In this scenario, both sin x and sin 2x are either positive or equal to zero. This means that the absolute value signs can be safely removed, and our equation simplifies to sin x = sin 2x. Using the double-angle identity, we can rewrite this as sin x = 2 sin x cos x. Now, let's rearrange the equation to bring all terms to one side: 2 sin x cos x - sin x = 0. We can then factor out sin x, giving us sin x (2 cos x - 1) = 0. This factored form tells us that either sin x = 0 or 2 cos x - 1 = 0. If sin x = 0, then x = nπ, where n is an integer. If 2 cos x - 1 = 0, then cos x = 1/2, which means x = ±π/3 + 2mπ, where m is an integer. But we must also ensure that in this case both sin x and sin 2x are non-negative. For x = nπ, sin x = 0 and sin 2x = 0, so these solutions are valid. For x = π/3 + 2mπ, sin x > 0 and sin 2x > 0, so these solutions are valid. For x = -π/3 + 2mπ, sin x < 0 and sin 2x < 0, so these solutions are not valid. Thus, the solutions in this case are x = nπ and x = π/3 + 2mπ. Congratulations, we solved the first case!

Case 2: sin x is non-negative and sin 2x is negative

In this case, sin x is either positive or zero, while sin 2x is negative. Therefore, our original equation transforms into sin x = -sin 2x. Using the double-angle identity, we rewrite this as sin x = -2 sin x cos x. Rearranging gives us 2 sin x cos x + sin x = 0. Factoring out sin x, we get sin x (2 cos x + 1) = 0. Thus, either sin x = 0 or 2 cos x + 1 = 0. If sin x = 0, then x = nπ, where n is an integer. If 2 cos x + 1 = 0, then cos x = -1/2, meaning x = ±2π/3 + 2mπ, where m is an integer. Now, let's check which of these solutions meet the case's conditions. For x = nπ, sin x = 0 and sin 2x = 0, which does not meet the requirements. For x = 2π/3 + 2mπ, sin x > 0 and sin 2x < 0, so these solutions are valid. For x = -2π/3 + 2mπ, sin x < 0 and sin 2x > 0, which do not meet the requirements. So the solutions in this case are x = 2π/3 + 2mπ.

Case 3: sin x is negative and sin 2x is non-negative

Here, sin x is negative, and sin 2x is either positive or zero. Our original equation becomes -sin x = sin 2x. Using the double-angle identity, we get -sin x = 2 sin x cos x. Rearranging the equation yields 2 sin x cos x + sin x = 0, which is the same as in Case 2. Therefore, either sin x = 0 or 2 cos x + 1 = 0. If sin x = 0, then x = nπ, where n is an integer. If 2 cos x + 1 = 0, then cos x = -1/2, which means x = ±2π/3 + 2mπ, where m is an integer. Again, we need to check if the solutions satisfy the conditions of this case. For x = nπ, sin x = 0 and sin 2x = 0, not valid. For x = 2π/3 + 2mπ, sin x > 0 and sin 2x < 0, not valid. For x = -2π/3 + 2mπ, sin x < 0 and sin 2x > 0, thus valid. The solutions are x = -2π/3 + 2mπ.

Case 4: sin x and sin 2x are both negative

In this final case, both sin x and sin 2x are negative. The equation transforms into -sin x = -sin 2x, or sin x = sin 2x, which is the same as in Case 1. Thus, we have sin x = 2 sin x cos x. Rearranging gives us sin x (2 cos x - 1) = 0, and therefore sin x = 0 or cos x = 1/2. If sin x = 0, then x = nπ. If cos x = 1/2, then x = ±π/3 + 2mπ. Now we check the conditions. For x = nπ, sin x = 0 and sin 2x = 0, which does not meet the conditions. For x = π/3 + 2mπ, sin x > 0 and sin 2x > 0, not valid. For x = -π/3 + 2mπ, sin x < 0 and sin 2x < 0, thus valid. The solutions are x = -π/3 + 2mπ.

Consolidating the Solutions: Putting it all Together

Alright, guys! We've successfully navigated through all four cases, and now it's time to consolidate our findings. After a careful analysis of each case, here's a summary of the solutions:

• From Case 1: x = nπ and x = π/3 + 2mπ.
• From Case 2: x = 2π/3 + 2mπ.
• From Case 3: x = -2π/3 + 2mπ.
• From Case 4: x = -π/3 + 2mπ.

Combining these, the general solution to the equation |sin x| = |sin 2x| is: x = nπ, x = ±π/3 + 2kπ and x = ±2π/3 + 2lπ, where n, k and l are integers. Therefore, we've found all possible values of x that satisfy the original equation. We've done it! We've meticulously worked through each case, applied our knowledge of trigonometric identities, and used logical reasoning to arrive at the solution.

Wrapping Up and Further Exploration

Congratulations on making it this far! You've successfully solved the absolute value trigonometric equation |sin x| = |sin 2x|. This problem demonstrates the power of case analysis in solving seemingly complex mathematical problems. By breaking down the problem into smaller, more manageable parts, we were able to find a comprehensive solution. Remember, the key to success in mathematics is practice and a willingness to explore. If you enjoyed this problem, I encourage you to try similar ones. Explore other trigonometric equations, experiment with different identities, and challenge yourself. The more you practice, the more comfortable you will become with these types of problems. Also, consider exploring more advanced topics in trigonometry, such as inverse trigonometric functions and trigonometric series. There's a whole world of mathematical wonders out there just waiting to be explored. Keep asking questions, keep learning, and most importantly, keep having fun with math! Thanks for joining me on this mathematical journey. Until next time, keep those mathematical muscles flexed and your curiosity alive!