Solving $2\sin^2 Θ - 3\sin Θ + 1 = 0$: A Step-by-Step Guide

Hey guys! Let's dive into solving the trigonometric equation $2 \sin^2 \theta - 3 \sin \theta + 1 = 0$. If you're scratching your head, don't worry! We're going to break it down step by step, so you'll be a pro in no time. Trigonometric equations might seem daunting at first, but they're actually quite manageable once you understand the underlying principles. This equation is a classic example of a quadratic trigonometric equation, and solving it involves a few key techniques. We'll explore these techniques in detail, making sure you grasp every concept along the way. So, grab your thinking caps, and let’s get started on this mathematical adventure! We’ll cover everything from the initial setup to the final solutions, ensuring you have a solid understanding of the process.

1. Recognizing the Quadratic Form

Okay, first things first. Recognizing the quadratic form is super important. You'll notice our equation, $2 \sin^2 \theta - 3 \sin \theta + 1 = 0$, looks a lot like a quadratic equation if we think of $\sin \theta$ as our variable. Think of it like $2x^2 - 3x + 1 = 0$. See the resemblance? This is our key to unlocking the solution. Identifying this quadratic structure allows us to use familiar algebraic techniques to solve for $\sin \theta$ first. This is a crucial step because it transforms a seemingly complex trigonometric problem into a more manageable algebraic one. By recognizing this form, we can apply methods such as factoring or the quadratic formula, which we'll discuss in detail later. So, always keep an eye out for this pattern when dealing with trigonometric equations! This approach simplifies the problem and makes it easier to find the solutions.

The Importance of Substitution

To make things even clearer, let’s use a substitution. Let $x = \sin \theta$. Now our equation transforms into $2x^2 - 3x + 1 = 0$. Much friendlier, right? This substitution simplifies the equation, making it easier to work with and visualize. It’s a common trick in mathematics to use substitutions to turn complex problems into simpler forms. By replacing $\sin \theta$ with $x$, we can now focus on solving a standard quadratic equation. This step is particularly helpful if you're more comfortable with algebraic manipulations than trigonometric ones. Once we find the values of $x$, we can then substitute back to find the values of $\theta$. This method not only simplifies the process but also reduces the chances of making mistakes along the way. So, remember the power of substitution! It's a fantastic tool in your mathematical toolkit.

2. Solving the Quadratic Equation

Now that we have $2x^2 - 3x + 1 = 0$, it’s time to solve this quadratic equation. There are a couple of ways we can tackle this: factoring or using the quadratic formula. Factoring is often the quicker route if you can spot the factors easily. The quadratic formula is a more general approach that always works, even when factoring isn't straightforward.

Factoring Method

Let's try factoring first. We need to find two numbers that multiply to $2 imes 1 = 2$ and add up to $-3$. Those numbers are $-2$ and $-1$. So, we can rewrite our equation as:

$2x^2 - 2x - x + 1 = 0$

Now, factor by grouping:

$2x(x - 1) - 1(x - 1) = 0$

$(2x - 1)(x - 1) = 0$

Setting each factor to zero gives us:

$2x - 1 = 0$ or $x - 1 = 0$

Solving for $x$, we get:

$x = \frac{1}{2}$ or $x = 1$

Quadratic Formula Method

If factoring isn't your jam, no worries! The quadratic formula is here to save the day. For an equation of the form $ax^2 + bx + c = 0$, the quadratic formula is:

$x = \frac{-b ${}$ \sqrt{b^2 - 4ac}}{2a}$

In our case, $a = 2$, $b = -3$, and $c = 1$. Plugging these values into the formula, we get:

$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(1)}}{2(2)}$

$x = \frac{3 \pm \sqrt{9 - 8}}{4}$

$x = \frac{3 \pm \sqrt{1}}{4}$

$x = \frac{3 \pm 1}{4}$

This gives us two solutions:

$x = \frac{3 + 1}{4} = 1$

$x = \frac{3 - 1}{4} = \frac{1}{2}$

See? We got the same solutions using both methods. It’s always good to have options! Understanding both factoring and the quadratic formula ensures you’re well-equipped to tackle any quadratic equation that comes your way. Choose the method you’re most comfortable with, or use both to double-check your work. The key is to be confident in your ability to find the solutions, no matter the approach you take.

3. Substituting Back and Solving for θ

Alright, we've found our values for $x$, but remember, we're actually looking for $\theta$. So, we need to substitute back $\sin \theta$ for $x$. This means we have two equations to solve:

$\sin \theta = \frac{1}{2}$

$\sin \theta = 1$

Let's tackle each of these separately.

Solving $\sin \theta = \frac{1}{2}$

To find the values of $\theta$ for which $\sin \theta = \frac{1}{2}$, we need to think about the unit circle. Do you remember your special angles? We know that $\sin \frac{\pi}{6} = \frac{1}{2}$. So, $\theta = \frac{\pi}{6}$ is one solution. But wait, there’s more!

Sine is also positive in the second quadrant. The reference angle in the second quadrant is $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$. So, $\theta = \frac{5\pi}{6}$ is another solution.

Since sine is periodic with a period of $2\pi$, we can add multiples of $2\pi$ to these solutions to get all possible solutions:

$\theta = \frac{\pi}{6} + 2n\pi$, where $n$ is an integer

$\theta = \frac{5\pi}{6} + 2n\pi$, where $n$ is an integer

Solving $\sin \theta = 1$

Now let's solve $\sin \theta = 1$. Again, think about the unit circle. At what angle is the sine equal to 1? That’s right, at $\theta = \frac{\pi}{2}$.

Since sine is 1 only at $\frac{\pi}{2}$ within the interval $[0, 2\pi)$, we just need to add multiples of $2\pi$ to get all solutions:

$\theta = \frac{\pi}{2} + 2n\pi$, where $n$ is an integer

This step of substituting back is crucial because it connects our algebraic solutions back to the original trigonometric context. Understanding the periodic nature of trigonometric functions is key to finding all possible solutions. Remember to consider all quadrants where the sine function has the required value. The unit circle is your best friend here, so make sure you're comfortable visualizing it and identifying the angles that correspond to specific sine values. By carefully considering the periodic nature and the unit circle, you can confidently find all solutions for $\theta$.

4. Summarizing the Solutions

Alright, we've done the heavy lifting! Let's summarize our solutions. We found three sets of solutions for the equation $2 \sin^2 \theta - 3 \sin \theta + 1 = 0$:

$\theta = \frac{\pi}{6} + 2n\pi$, where $n$ is an integer

$\theta = \frac{5\pi}{6} + 2n\pi$, where $n$ is an integer

$\theta = \frac{\pi}{2} + 2n\pi$, where $n$ is an integer

These are all the angles that satisfy our original equation. How cool is that? Summarizing the solutions is a critical step in any mathematical problem. It ensures that you have a clear understanding of all possible answers and that you haven't missed any. In this case, we've identified three distinct sets of solutions, each representing an infinite number of angles due to the periodic nature of the sine function. By expressing the solutions in this general form, we capture all possibilities. This final step solidifies your understanding of the problem and your ability to find comprehensive solutions.

Checking Your Answers

If you want to be extra sure (and you should!), you can plug these solutions back into the original equation to check that they work. This is a great way to catch any mistakes and build your confidence in your answers. Verifying your solutions is a best practice in mathematics, especially when dealing with complex equations. By substituting the solutions back into the original equation, you can confirm that they satisfy the equation and that no algebraic errors were made during the solving process. This step not only validates your answers but also reinforces your understanding of the problem-solving process. So, always take the time to check your work; it's a valuable investment in your mathematical accuracy and confidence.

Conclusion

So there you have it! We've successfully solved the trigonometric equation $2 \sin^2 \theta - 3 \sin \theta + 1 = 0$. We walked through recognizing the quadratic form, solving the quadratic equation, substituting back to find $\theta$, and summarizing our solutions. You guys are rockstars! Solving trigonometric equations is a fundamental skill in mathematics, with applications in various fields such as physics, engineering, and computer science. By mastering these techniques, you're not just solving equations; you're building a foundation for more advanced mathematical concepts. Keep practicing, and you'll become even more proficient at tackling these types of problems. Remember, the key is to break down complex problems into manageable steps, and you'll be solving equations like a pro in no time!

Keep practicing, and you'll become a master of trigonometric equations in no time. Happy solving!