Trigonometry Solutions: Mastering Equations Step-by-Step

Hey everyone! Today, we're diving into the world of trigonometry and tackling some equations. Don't worry, it's not as scary as it sounds! We'll break down each problem step-by-step, making sure you understand the 'why' behind the 'how.' So grab your pencils, and let's get started. We'll be solving for $\theta$ in a bunch of different scenarios. Ready to become trigonometry masters? Let's go!

18. Solving $(\sin \theta + 1)(\sin \theta - 1) = 0$

Alright, guys, let's kick things off with our first equation: $(\sin \theta + 1)(\sin \theta - 1) = 0$. This one is all about recognizing that when the product of two terms equals zero, one or both of those terms must be zero. This is a fundamental concept in algebra, and it's super helpful here. So, we'll set each factor equal to zero and solve for $\theta$. Firstly, we deal with the term $(\sin \theta + 1) = 0$. Subtracting 1 from both sides, we get $\sin \theta = -1$. Now, we need to think: at what angle(s) is the sine function equal to -1? Remember that the sine function corresponds to the y-coordinate on the unit circle. The y-coordinate is -1 at $\theta = \frac{3\pi}{2} + 2n\pi$, where 'n' is any integer. This represents all the coterminal angles. Next up, we have $(\sin \theta - 1) = 0$. Adding 1 to both sides, we get $\sin \theta = 1$. Again, let's think: when is the sine function equal to 1? This happens when the y-coordinate is 1 on the unit circle. This occurs at $\theta = \frac{\pi}{2} + 2n\pi$, where 'n' is any integer. So, our solutions for $\theta$ are $\frac{\pi}{2} + 2n\pi$ and $\frac{3\pi}{2} + 2n\pi$. Basically, any angle that gives us a y-coordinate of 1 or -1 on the unit circle satisfies the equation. It's like finding all the points where the sine wave touches either its peak or its trough.

To summarize this problem, we've used the zero-product property to break down the equation into simpler parts. We identified the angles where the sine function takes on the values of 1 and -1. The understanding of the unit circle and the behavior of the sine function is crucial here. Always remember that trigonometric functions are cyclical, and solutions repeat at regular intervals. Keep this in mind when you are tackling these problems. The solutions represent specific points on the unit circle, emphasizing the periodic nature of trigonometric functions. The method is straightforward: set each factor to zero, isolate the trigonometric function (in this case, sine), and identify the angles where the function equals the resulting value. Practice helps solidify these concepts, making solving trigonometric equations much more manageable.

19. Solving $(\tan \theta - 1)(\sec \theta - 1) = 0$

Alright, let's tackle the equation $(\tan \theta - 1)(\sec \theta - 1) = 0$. Just like before, we're dealing with the zero-product property, which means either $(\tan \theta - 1) = 0$ or $(\sec \theta - 1) = 0$, or both. Let's start with the first case: $(\tan \theta - 1) = 0$. Adding 1 to both sides, we get $\tan \theta = 1$. Now, remember that $\tan \theta$ is the ratio of $\sin \theta$ to $\cos \theta$. So, we're looking for angles where $\sin \theta$ and $\cos \theta$ have the same value (since their ratio is 1). This happens at $\theta = \frac{\pi}{4} + n\pi$, where 'n' is any integer. These are angles in the first and third quadrants of the unit circle, where the x and y values (corresponding to cosine and sine, respectively) are equal in magnitude and have the same sign. In the second case, $(\sec \theta - 1) = 0$. Adding 1 to both sides gives us $\sec \theta = 1$. Recall that $\sec \theta$ is the reciprocal of $\cos \theta$. So, $\sec \theta = 1$ when $\cos \theta = 1$. This occurs at $\theta = 2n\pi$, where 'n' is any integer. These are angles that fall on the positive x-axis of the unit circle, at intervals of 2$\pi$ radians. The core idea here is understanding the definitions of tangent and secant in relation to sine and cosine. Recognizing the unit circle and knowing the values of trigonometric functions at key angles makes this process much smoother. The solutions highlight where these trigonometric functions take on specific values, revealing the geometric relationships of the unit circle. The crucial thing is converting from $\tan \theta$ and $\sec \theta$ to the more familiar $\sin \theta$ and $\cos \theta$ to make the problem easier to solve. Always remember the definitions of your trigonometric functions. Use your knowledge of the unit circle to see where they equal certain values. It's all about practice and building familiarity with the properties of these functions.

20. Solving $2\cos^2 \theta + \cos \theta = 0$

Okay, let's get into the equation $2\cos^2 \theta + \cos \theta = 0$. This one doesn't involve the zero-product property directly, but we can factor to solve it. Notice that we can factor out a $\cos \theta$ from both terms on the left side of the equation. So, we rewrite it as $\cos \theta(2\cos \theta + 1) = 0$. Now, we can apply the zero-product property! This means either $\cos \theta = 0$ or $2\cos \theta + 1 = 0$. Let's start with the first case: $\cos \theta = 0$. The cosine function corresponds to the x-coordinate on the unit circle. So, we're looking for angles where the x-coordinate is 0. This happens at $\theta = \frac{\pi}{2} + n\pi$, where 'n' is any integer. These are angles located on the positive and negative y-axis. Next, we have $2\cos \theta + 1 = 0$. Subtracting 1 from both sides, we get $2\cos \theta = -1$. Dividing both sides by 2, we have $\cos \theta = -\frac{1}{2}$. We know that $\cos \theta$ is negative in the second and third quadrants. So, we're looking for angles where the x-coordinate is -1/2. This happens at $\theta = \frac{2\pi}{3} + 2n\pi$ and $\theta = \frac{4\pi}{3} + 2n\pi$, where 'n' is any integer. Remember, the angle $\frac{2\pi}{3}$ is in the second quadrant, and $\frac{4\pi}{3}$ is in the third quadrant. This problem is solved using a combination of factoring and using your unit circle knowledge. First, we factored the equation, which gave us two simpler expressions. Then, we solved each expression by finding the angles where the cosine function takes on specific values. Remember, practice is key! Familiarize yourself with the unit circle, and the values of sine and cosine at various angles. Always be on the lookout for opportunities to factor. This often simplifies the equation and makes it easier to solve. Factoring allows us to break a complex equation into smaller, more manageable pieces, applying what we already know about the unit circle. Always make sure you understand the basics of the trigonometric functions. This knowledge will serve you well in various problem-solving scenarios.

21. Solving $2\sin^2 \theta + \sin \theta - 1 = 0$

Alright, let's tackle $2\sin^2 \theta + \sin \theta - 1 = 0$. This equation looks like a quadratic equation in disguise. We can solve this by factoring. Think of $\sin \theta$ as 'x' for a moment. The equation becomes $2x^2 + x - 1 = 0$. Can we factor this? Absolutely! It factors into $(2\sin \theta - 1)(\sin \theta + 1) = 0$. Now, we can apply the zero-product property. Either $2\sin \theta - 1 = 0$ or $\sin \theta + 1 = 0$. Let's solve the first case: $2\sin \theta - 1 = 0$. Adding 1 to both sides gives us $2\sin \theta = 1$. Dividing both sides by 2, we get $\sin \theta = \frac{1}{2}$. The sine function is positive in the first and second quadrants. We know that $\sin \theta = \frac{1}{2}$ at $\theta = \frac{\pi}{6} + 2n\pi$ and $\theta = \frac{5\pi}{6} + 2n\pi$, where 'n' is any integer. Next, let's solve the second case: $\sin \theta + 1 = 0$. Subtracting 1 from both sides gives us $\sin \theta = -1$. We've seen this before! This happens at $\theta = \frac{3\pi}{2} + 2n\pi$, where 'n' is any integer. Remember, it's all about recognizing patterns and understanding how trigonometric functions behave. This problem demonstrates the power of factoring to solve trigonometric equations. By recognizing the equation as a quadratic in terms of $\sin \theta$, we were able to factor it and apply the zero-product property. Always be on the lookout for these kinds of patterns. This method highlights the interplay between algebra and trigonometry, showcasing how factoring techniques can be leveraged to simplify and solve complex equations. This problem requires a solid understanding of factoring quadratic expressions and the unit circle. Practicing such problems helps to reinforce the relationship between the algebraic structure and the trigonometric functions' geometric interpretations. The factoring step transforms a complex equation into manageable expressions, simplifying the overall solving process.

22. Solving $\sin^2 \theta - \cos^2 \theta = 1 + \cos \theta$

Let's get into the equation $\sin^2 \theta - \cos^2 \theta = 1 + \cos \theta$. This one looks a little different, but we can simplify it using a trigonometric identity. Remember the Pythagorean identity: $\sin^2 \theta + \cos^2 \theta = 1$? We can rearrange this to get $\sin^2 \theta = 1 - \cos^2 \theta$. Now, substitute this into our original equation: $(1 - \cos^2 \theta) - \cos^2 \theta = 1 + \cos \theta$. Simplifying, we get $1 - 2\cos^2 \theta = 1 + \cos \theta$. Now, let's move everything to one side of the equation: $0 = 2\cos^2 \theta + \cos \theta$. Wait a minute! We've seen this before in problem 20! We can factor out a $\cos \theta$: $\cos \theta(2\cos \theta + 1) = 0$. Just like before, this means either $\cos \theta = 0$ or $2\cos \theta + 1 = 0$. We already know the solutions to these from problem 20. When $\cos \theta = 0$, we have $\theta = \frac{\pi}{2} + n\pi$, where 'n' is any integer. When $2\cos \theta + 1 = 0$, we have $\cos \theta = -\frac{1}{2}$, which gives us $\theta = \frac{2\pi}{3} + 2n\pi$ and $\theta = \frac{4\pi}{3} + 2n\pi$, where 'n' is any integer. This problem showcases the importance of knowing and applying trigonometric identities. Specifically, we used the Pythagorean identity to rewrite the equation in terms of a single trigonometric function (cosine). This is a common technique used to simplify trigonometric equations. Once we had the equation in terms of cosine, we were able to factor and solve it, just like we did in problem 20. The ability to manipulate trigonometric equations using identities is a crucial skill. Mastering identities allows us to transform complex equations into simpler forms, often revealing hidden patterns. Always look for opportunities to simplify the equation using fundamental trigonometric identities. The application of identities transformed the problem into a more familiar format that we already knew how to solve, highlighting the power of algebraic manipulation within the context of trigonometry. Regular practice and a strong understanding of fundamental identities is key to success.

23. Solving $\sin^2 \theta = 6\cos \theta + 6$

Let's wrap things up with $\sin^2 \theta = 6\cos \theta + 6$. Again, we'll use a trigonometric identity. Replace $\sin^2 \theta$ with $1 - \cos^2 \theta$ (from the Pythagorean identity). This gives us $1 - \cos^2 \theta = 6\cos \theta + 6$. Rearranging, we get $\cos^2 \theta + 6\cos \theta + 5 = 0$. This is another quadratic equation! Let's solve it by factoring: $(\cos \theta + 5)(\cos \theta + 1) = 0$. Now, by the zero-product property, either $\cos \theta + 5 = 0$ or $\cos \theta + 1 = 0$. In the first case, $\cos \theta = -5$. But wait! The range of the cosine function is [-1, 1]. So, there's no solution here. In the second case, $\cos \theta = -1$. This happens at $\theta = \pi + 2n\pi$, where 'n' is any integer. This final example highlights a crucial aspect of trigonometry: always consider the range of the trigonometric functions. While the equation initially looked like a standard quadratic, the range of the cosine function meant that one of the solutions was impossible. This underscores the need to be aware of the properties of the functions you're working with. This problem is a great reminder that not all solutions are valid. The range of trigonometric functions places limitations on the possible solutions, and we must always check for consistency. The critical step here was the identification of the trigonometric identity. Always be careful to double-check that your answers fit the range of the trigonometric function. Mastering this requires a deep understanding of trigonometric identities, algebraic manipulation, and the properties of trigonometric functions. Practicing problems like this helps develop critical thinking skills, which are essential for tackling more complex equations. Congratulations, you've reached the end! Keep up the great work. Remember, practice is key to mastering trigonometry. The more you work through these problems, the more confident you'll become. Keep practicing and exploring, and you'll become a trigonometry whiz in no time!