Solving Sin(2x) = Sin(x) Find Solutions Over [0, 2π)

Hey there, math enthusiasts! Today, we're diving into a trigonometric equation that might seem a bit tricky at first, but don't worry, we'll break it down step by step. We're going to find all the solutions to the equation $\sin(2x) = \sin(x)$ within the interval $[0, 2\pi)$. So, let's put on our thinking caps and get started!

Understanding the Problem

Before we jump into solving, let's make sure we understand what the question is asking. We need to find all the values of x that make the equation $\sin(2x) = \sin(x)$ true, but only the values that fall between 0 and $2\pi$ (including 0 but not including $2\pi$). This interval represents one full revolution around the unit circle, which is why it's a common range to consider in trigonometry problems.

Trigonometric equations can sometimes feel like puzzles, but the key is to use trigonometric identities and algebraic techniques to simplify them. In this case, the double-angle identity for sine will be our best friend. Remember, identities are equations that are always true, no matter what value you plug in for the variable.

So, what's our plan of attack? First, we'll use the double-angle identity to rewrite $\sin(2x)$. Then, we'll rearrange the equation to get everything on one side, and hopefully, we'll be able to factor it. Factoring is a super useful tool in solving equations because it allows us to break a complex problem into simpler ones. Once we've factored, we can set each factor equal to zero and solve for x. Finally, we'll check our solutions to make sure they fall within the given interval. Sounds like a plan? Let's do it!

Applying the Double-Angle Identity

Our main keyword here is the double-angle identity for sine. This identity states that $\sin(2x) = 2\sin(x)\cos(x)$. It's a crucial tool for simplifying expressions and solving equations involving sine and cosine. Think of it as a recipe for turning a $\sin(2x)$ term into something more manageable. Why is this helpful? Well, it allows us to rewrite our original equation in terms of $\sin(x)$ and $\cos(x)$, which will make it easier to manipulate.

So, let's replace $\sin(2x)$ in our equation with $2\sin(x)\cos(x)$. Our equation now looks like this: $2\sin(x)\cos(x) = \sin(x)$. See how much simpler it's becoming? Now we have a mix of sine and cosine terms, but they're all related to x, not 2x. This is a big step forward!

The next step is to rearrange the equation. We want to get all the terms on one side so that we have zero on the other side. This is a common strategy in solving equations because it sets us up for factoring. So, let's subtract $\sin(x)$ from both sides of the equation. This gives us: $2\sin(x)\cos(x) - \sin(x) = 0$. Great! Now we're looking at an equation that's ripe for factoring.

Factoring and Solving

Now comes the fun part: factoring. Look closely at the left side of the equation: $2\sin(x)\cos(x) - \sin(x) = 0$. Do you see a common factor? That's right, both terms have a $\sin(x)$ in them! Factoring out $\sin(x)$ is like pulling out a common ingredient in a recipe. It simplifies the expression and makes it easier to handle.

When we factor out $\sin(x)$, we get: $\sin(x)(2\cos(x) - 1) = 0$. This is a huge breakthrough! We've transformed our trigonometric equation into a product of two factors that equals zero. And you know what that means, right? It means that at least one of the factors must be zero.

This is the zero-product property in action. It's a fundamental principle in algebra that says if a * b = 0, then either a = 0 or b = 0 (or both). We're going to use this property to split our equation into two simpler equations: $\sin(x) = 0$ and $2\cos(x) - 1 = 0$. Now we have two separate trigonometric equations to solve, which is much easier than tackling the original one.

Let's start with the first equation: $\sin(x) = 0$. Think about the unit circle. Where on the unit circle is the y-coordinate (which represents the sine value) equal to zero? There are two places: at 0 radians and at $\pi$ radians. So, two potential solutions are x = 0 and x = $\pi$. But remember, we're working in the interval $[0, 2\pi)$, so we need to consider all possible solutions within that range.

Now let's tackle the second equation: $2\cos(x) - 1 = 0$. This one requires a little more algebraic manipulation. First, we add 1 to both sides: $2\cos(x) = 1$. Then, we divide both sides by 2: $\cos(x) = \frac{1}{2}$. Again, think about the unit circle. Where is the x-coordinate (which represents the cosine value) equal to $\frac{1}{2}$? There are two places: in the first quadrant and in the fourth quadrant.

In the first quadrant, the angle whose cosine is $\frac{1}{2}$ is $\frac{\pi}{3}$ radians. In the fourth quadrant, the angle whose cosine is $\frac{1}{2}$ is $\frac{5\pi}{3}$ radians. So, we have two more potential solutions: x = $\frac{\pi}{3}$ and x = $\frac{5\pi}{3}$.

Checking the Solutions

Okay, we've found four potential solutions: 0, $\pi$, $\frac{\pi}{3}$, and $\frac{5\pi}{3}$. But before we declare victory, we need to check that these solutions actually work in the original equation and that they fall within the interval $[0, 2\pi)$. This is a crucial step because sometimes we can introduce extraneous solutions when we solve equations.

Let's start with x = 0. Plugging this into the original equation, we get $\sin(2 * 0) = \sin(0)$, which simplifies to $\sin(0) = \sin(0)$. Since $\sin(0) = 0$, this solution works! And 0 is definitely within our interval.

Next, let's check x = $\pi$. Plugging this in, we get $\sin(2 * \pi) = \sin(\pi)$, which simplifies to $\sin(2\pi) = \sin(\pi)$. Since $\sin(2\pi) = 0$ and $\sin(\pi) = 0$, this solution also works! And $\pi$ is within our interval.

Now, let's check x = $\frac{\pi}{3}$. Plugging this in, we get $\sin(2 * \frac{\pi}{3}) = \sin(\frac{\pi}{3})$, which simplifies to $\sin(\frac{2\pi}{3}) = \sin(\frac{\pi}{3})$. We know that $\sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$, so this solution works too! And $\frac{\pi}{3}$ is within our interval.

Finally, let's check x = $\frac{5\pi}{3}$. Plugging this in, we get $\sin(2 * \frac{5\pi}{3}) = \sin(\frac{5\pi}{3})$, which simplifies to $\sin(\frac{10\pi}{3}) = \sin(\frac{5\pi}{3})$. Since $\frac{10\pi}{3}$ is more than $2\pi$, we need to find its coterminal angle within the interval $[0, 2\pi)$. Subtracting $2\pi$ (or $\frac{6\pi}{3}$) from $\frac{10\pi}{3}$ gives us $\frac{4\pi}{3}$. So, we have $\sin(\frac{4\pi}{3}) = \sin(\frac{5\pi}{3})$. We know that $\sin(\frac{4\pi}{3}) = -\frac{\sqrt{3}}{2}$ and $\sin(\frac{5\pi}{3}) = -\frac{\sqrt{3}}{2}$, so this solution also works! And $\frac{5\pi}{3}$ is within our interval.

The Solutions

We've done it! We've found all the solutions to the equation $\sin(2x) = \sin(x)$ within the interval $[0, 2\pi)$. Our solutions are: x = 0, x = $\pi$, x = $\frac{\pi}{3}$, and x = $\frac{5\pi}{3}$.

To summarize, we used the double-angle identity to rewrite the equation, factored to simplify it, applied the zero-product property to break it into smaller equations, solved those equations using our knowledge of the unit circle, and then checked our solutions to make sure they worked. Phew! That's a lot, but you did great!

Remember, solving trigonometric equations is like detective work. You need to use all the tools at your disposal—identities, algebraic techniques, and the unit circle—to uncover the hidden solutions. And always, always check your work!

Final Answer

So, the solutions to the equation $\sin(2x) = \sin(x)$ over the interval $[0, 2\pi)$ are:

• 0
• $\frac{\pi}{3}$
• $\pi$
• $\frac{5\pi}{3}$

That's all for today, folks! Keep practicing, and you'll become a trigonometry master in no time! If you have any questions or want to explore more trigonometric equations, feel free to ask. Happy solving!