Solving Trigonometric Equations: A Step-by-Step Guide

Hey math enthusiasts! Today, we're diving into the world of trigonometry and tackling a specific equation: $4 \sin^2 \theta + 4 \sin \theta + 1 = 0$. Our goal is to find all the values of $\theta$ that satisfy this equation, but with a catch – we're only looking within the range of $0 \leq \theta < 2\pi$. Don't worry if that sounds intimidating; we'll break it down step by step to make it super clear. This is a common type of problem you might encounter in your math journey, so understanding it will give you a solid foundation for more complex trigonometric challenges. Let's get started!

Understanding the Problem and Initial Steps

First, let's take a closer look at the equation: $4 \sin^2 \theta + 4 \sin \theta + 1 = 0$. At first glance, it might seem a bit complicated, but it actually has a nice, familiar structure. Notice how it involves $\sin \theta$ and its square. This is a big hint! This equation is a quadratic equation disguised in trigonometric clothing. Our primary objective is to solve this equation for $\theta$ within the specified range. The core concept here is to treat $\sin \theta$ as a variable. This makes it easier to recognize the underlying quadratic form. We will manipulate the equation using algebraic techniques and trigonometric identities, and then solve for the variable that we have. We should not be afraid of the problem. Remember to break down the problem into smaller and more manageable tasks. Always check the unit circle to see how the equations works. Now, let's proceed with solving for $\theta$.

To make things simpler, let's introduce a substitution. Let $x = \sin \theta$. Now, our equation becomes: $4x^2 + 4x + 1 = 0$. Recognize this equation? It's a standard quadratic equation. This substitution simplifies the problem, making it easier to solve using techniques you're already familiar with. By replacing $\sin \theta$ with $x$, we transform the original equation into a form that's much more manageable and easier to analyze. In the next step, we'll solve this quadratic equation to find the value of $x$, and then we can get back to $\theta$. This is all about taking a complex problem and breaking it into easier to deal with parts. The substitution method is crucial in solving more complex equations.

Solving the Quadratic Equation

Now that we've got our quadratic equation, $4x^2 + 4x + 1 = 0$, it's time to solve it. Fortunately, this particular quadratic equation is a perfect square trinomial. This means it can be factored very nicely. The equation fits the pattern $(ax + b)^2$. Let us rewrite the equation in this format. We need to find values $a$ and $b$ so that the following equation can be satisfied: $(ax + b)^2 = 4x^2 + 4x + 1$. Expanding the left side of the equation yields: $a^2x^2 + 2abx + b^2 = 4x^2 + 4x + 1$. Now, compare the coefficients. We can see that $a^2 = 4$, $2ab = 4$, and $b^2 = 1$. It looks like $a=2$ and $b=1$ will satisfy all the equations. Now, the equation can be written as $(2x + 1)^2 = 0$. This step is a critical part of the solution process as it significantly simplifies the problem.

So, we can rewrite the equation as $(2x + 1)^2 = 0$. To solve for $x$, we take the square root of both sides, which gives us $2x + 1 = 0$. Now, a simple algebraic manipulation will yield the solution. Solving for $x$ in $2x + 1 = 0$, we subtract 1 from both sides to get $2x = -1$, and then divide both sides by 2 to get $x = -\frac{1}{2}$. This value of $x$ represents the sine of $\theta$. Now that we've solved for $x$, it's time to bring back our original variable, $\theta$, using the substitution we made earlier.

Finding the Values of θ

Remember that we initially set $x = \sin \theta$. Now that we know $x = -\frac{1}{2}$, we can substitute it back into our original equation. This means we need to find the values of $\theta$ for which $\sin \theta = -\frac{1}{2}$. This part involves using your knowledge of the unit circle or the sine function's properties. In the unit circle, the sine value corresponds to the y-coordinate. So, we're looking for angles where the y-coordinate is $-\frac{1}{2}$. The unit circle is a powerful tool for visualizing trigonometric functions and their values at various angles. Think about where the sine function has negative values. Sine is negative in the third and fourth quadrants. Now, we're looking for angles in the range of $0 \leq \theta < 2\pi$ where $\sin \theta = -\frac{1}{2}$. These angles will be in the third and fourth quadrants.

The angles where $\sin \theta = -\frac{1}{2}$ are $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$. To find these angles, you can use your knowledge of special triangles (like the 30-60-90 triangle) or a calculator. Recall that $\sin(\frac{\pi}{6}) = \frac{1}{2}$. Since we are looking for negative values, we need to find the angles in the third and fourth quadrants where the reference angle is $\frac{\pi}{6}$. The angle in the third quadrant is $\pi + \frac{\pi}{6} = \frac{7\pi}{6}$, and the angle in the fourth quadrant is $2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$.

Therefore, the solutions to the equation $4 \sin^2 \theta + 4 \sin \theta + 1 = 0$ for $0 \leq \theta < 2\pi$ are $\theta = \frac{7\pi}{6}$ and $\theta = \frac{11\pi}{6}$. Remember to always check your answers to make sure they fit within the initially given constraints.

Conclusion and Key Takeaways

Congratulations! You've successfully solved the trigonometric equation. We started with $4 \sin^2 \theta + 4 \sin \theta + 1 = 0$ and, by using substitution and our understanding of quadratic equations and trigonometric functions, we found that $\theta = \frac{7\pi}{6}$ and $\theta = \frac{11\pi}{6}$. This problem nicely illustrates how different areas of math can be connected and how a little bit of algebraic manipulation can make a complex problem much more approachable. The key takeaways from this problem are:

• Recognize the patterns: Look for ways to simplify the equation by recognizing familiar structures, like the quadratic form. This lets you apply techniques you already know.
• Use substitution: Introduce a substitution to simplify the equation, making it easier to solve. Don't forget to substitute back at the end!
• Utilize the unit circle: The unit circle is your friend! It's a great visual tool to understand the values of sine and cosine at different angles.
• Understand trigonometric identities: Knowing basic trigonometric identities can help you simplify and solve trigonometric equations more efficiently.

By following these steps, you can confidently approach and solve other trigonometric equations. Practice is key, so try solving similar problems to reinforce your understanding. Keep exploring the world of mathematics, and don't hesitate to ask questions. Happy solving!

For further reading and more practice problems, you can visit Khan Academy's Trigonometry Section. This website offers a wealth of resources, including video tutorials and practice exercises, that can help you master trigonometric concepts and improve your problem-solving skills.