Solving Trigonometric Equations: Sin(2x) - Sin(x) = 0

Hey there, math enthusiasts! Today, we're diving into the fascinating world of trigonometric equations. Specifically, we'll be tackling the problem: Let $\sin (2 x)-\sin (x)=0$, where $0 \leq x<2 \pi$. What are the possible solutions for $x$? This kind of problem is super common in pre-calculus and calculus, and understanding how to solve it will unlock a whole new level of mathematical prowess for you. We'll break down the steps, explain the reasoning behind them, and make sure you're comfortable with the process. So, grab your calculators, your notebooks, and let's get ready to explore the unit circle and its incredible properties!

Understanding the Equation: $\sin (2 x)-\sin (x)=0$

Our journey begins with the equation $\sin (2 x)-\sin (x)=0$. The first thing you'll notice is the presence of $\sin(2x)$. This is a double-angle identity, and recognizing it is key to simplifying our problem. The double-angle identity for sine is $\sin(2x) = 2\sin(x)\cos(x)$. By substituting this into our original equation, we transform it into something much more manageable. So, let's rewrite the equation using this identity:

$2\sin(x)\cos(x) - \sin(x) = 0$

Now, we have an equation that involves both $\sin(x)$ and $\cos(x)$. The next crucial step in solving many algebraic and trigonometric equations is factoring. See that $\sin(x)$ term appearing in both parts of our expression? That means we can factor it out! This is a powerful technique that will help us isolate the variables and find our solutions. Let's factor out $\sin(x)$:

$\sin(x)(2\cos(x) - 1) = 0$

This factored form is a game-changer. The principle of zero products states that if the product of two or more factors is zero, then at least one of the factors must be zero. This means we can set each of our factors equal to zero and solve for $x$ independently. This will give us two separate, simpler equations to work with:

1. $\sin(x) = 0$
2. $2\cos(x) - 1 = 0$

We're well on our way to finding all the possible values of $x$ within the given interval $0 \leq x<2 \pi$. Remember, understanding these fundamental trigonometric identities and algebraic manipulation techniques is the bedrock of solving more complex problems. Let's keep going!

Solving for $x$ when $\sin(x) = 0$

Our first equation is $\sin(x) = 0$. We need to find all the values of $x$ in the interval $[0, 2\pi)$ for which the sine function equals zero. Think about the unit circle. The sine of an angle corresponds to the y-coordinate of the point where the terminal side of the angle intersects the unit circle. So, we're looking for angles where the y-coordinate is 0.

On the unit circle, the y-coordinate is zero at two specific points: when the angle is $0$ radians (or $0^\circ$) and when the angle is $\pi$ radians (or $180^\circ$). Since our interval is $0 \leq x<2 \pi$, both $x=0$ and $x=\pi$ are valid solutions.

It's important to be precise with the interval. The interval $0 \leq x<2 \pi$ includes $0$ but excludes $2\pi$. If we were to consider $2\pi$, it would represent the same position on the unit circle as $0$, and since $0$ is already included, $2\pi$ wouldn't add a new unique solution in this context. So, from $\sin(x) = 0$, we get two solutions: $x=0$ and $x=\pi$.

These are the first set of solutions we've found. Keep these in mind as we move on to the second part of our factored equation. The unit circle is your best friend when visualizing these trigonometric values, so don't hesitate to sketch one out if it helps you to see where these values come from. It’s all about connecting the abstract concept of sine to concrete points on a geometric figure.

Solving for $x$ when $2\cos(x) - 1 = 0$

Now, let's tackle the second equation we derived from our factoring: $2\cos(x) - 1 = 0$. To solve for $x$, we first need to isolate $\cos(x)$. We can do this by adding 1 to both sides of the equation, which gives us $2\cos(x) = 1$. Then, we divide both sides by 2 to get $\cos(x) = \frac{1}{2}$.

We are looking for angles $x$ in the interval $[0, 2\pi)$ where the cosine function equals $\frac{1}{2}$. Remember, the cosine of an angle corresponds to the x-coordinate of the point where the terminal side of the angle intersects the unit circle. So, we're searching for angles where the x-coordinate is $\frac{1}{2}$.

There are two primary angles in the interval $[0, 2\pi)$ that have a cosine of $\frac{1}{2}$. The first is in the first quadrant, and the second is in the fourth quadrant. The reference angle for which $\cos(\theta) = \frac{1}{2}$ is $\frac{\pi}{3}$ (or $60^\circ$).

• In the first quadrant, the angle is simply the reference angle itself. So, one solution is $x = \frac{\pi}{3}$. This is because $\cos(\frac{\pi}{3}) = \frac{1}{2}$.

• In the fourth quadrant, the angle can be found by subtracting the reference angle from $2\pi$. So, the other solution is $x = 2\pi - \frac{\pi}{3} = \frac{6\pi}{3} - \frac{\pi}{3} = \frac{5\pi}{3}$. This is because $\cos(\frac{5\pi}{3}) = \frac{1}{2}$.

Both of these angles, $\frac{\pi}{3}$ and $\frac{5\pi}{3}$, fall within our specified interval of $0 \leq x<2 \pi$. Therefore, from the equation $\cos(x) = \frac{1}{2}$, we have found two more solutions: $x = \frac{\pi}{3}$ and $x = \frac{5\pi}{3}$. Keep these noted as we compile our final answer.

Consolidating All Solutions

We have successfully solved both parts of our factored trigonometric equation. Let's bring all the solutions together and ensure they are within the specified interval $0 \leq x<2 \pi$.

From $\sin(x) = 0$, we found the solutions:

• $x = 0$
• $x = \pi$

From $2\cos(x) - 1 = 0$ (which simplified to $\cos(x) = \frac{1}{2}$), we found the solutions:

• $x = \frac{\pi}{3}$
• $x = \frac{5\pi}{3}$

Now, we just need to combine these unique solutions into a single set. The possible values for $x$ that satisfy the original equation $\sin(2x) - \sin(x) = 0$ in the interval $0 \leq x<2 \pi$ are:

$x \in \left\{0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}\right\}$

These are all the distinct solutions within the given range. It's always a good practice to double-check your answers by plugging them back into the original equation, though for this structured problem, our steps should ensure accuracy. Remember, the key was using the double-angle identity, factoring, and understanding the unit circle for sine and cosine values. This systematic approach is what makes solving these kinds of problems satisfying and achievable.

Final Thoughts on Solving Trigonometric Equations

Solving trigonometric equations like $\sin(2x) - \sin(x) = 0$ is a fundamental skill in mathematics, particularly for those pursuing STEM fields. The process we followed – utilizing trigonometric identities, employing algebraic techniques like factoring, and understanding the behavior of trigonometric functions on the unit circle – is a common thread in many advanced mathematical problems. The interval restriction $0 \leq x<2 \pi$ is also crucial, as it defines the domain within which we seek our solutions, preventing an infinite number of answers due to the periodic nature of sine and cosine.

It's worth reiterating the importance of recognizing common identities. The double-angle formula for sine, $\sin(2x) = 2\sin(x)\cos(x)$, was instrumental here. Without it, the problem would be significantly harder to approach. Similarly, the ability to factor an expression, $\sin(x)(2\cos(x) - 1) = 0$, allows us to break down a complex equation into simpler, solvable parts. Each part, $\sin(x)=0$ and $2\cos(x)-1=0$, represents a fundamental trigonometric equation that can be solved by recalling or deriving values from the unit circle.

When $\sin(x)=0$, we are looking for angles where the y-coordinate on the unit circle is zero. These are the angles lying on the x-axis, which are $0$ and $\pi$ within our interval. When $\cos(x)=\frac{1}{2}$, we are looking for angles where the x-coordinate on the unit circle is $\frac{1}{2}$. This occurs in the first and fourth quadrants, corresponding to angles $\frac{\pi}{3}$ and $\frac{5\pi}{3}$. The combination of these individual solutions gives us the complete set for the original equation.

For further exploration into trigonometry and its applications, you can check out resources like Khan Academy's Trigonometry Section. They offer comprehensive lessons, practice exercises, and video explanations that can deepen your understanding of these concepts. Mastering these skills will not only help you in your current studies but also equip you with valuable tools for future mathematical endeavors.