Solving Trigonometric Equations: A Deep Dive

In the realm of mathematics, solving trigonometric equations is a fundamental skill that unlocks a deeper understanding of periodic functions and their applications. Today, we're going to tackle a specific problem: solving the equation $-8 \sqrt{2} \csc (\theta)+6=-10$ for $\theta$ on the interval $[0, 2\pi)$. This might look a bit intimidating at first glance, with the cosecant function and the radical, but don't worry! We'll break it down step-by-step, making sure you understand each part of the process. Our goal is to find all the values of $\theta$ within a single rotation around the unit circle (from 0 up to, but not including, $2\pi$) that satisfy this equation. This involves isolating the trigonometric function, understanding its reciprocal relationship, and then using our knowledge of the unit circle to identify the correct angles. We'll also touch upon why restricting the interval is important in trigonometric equations, as many have infinitely many solutions without such a constraint. So, grab your notebooks, and let's embark on this mathematical journey together!

Isolating the Cosecant Function

The first crucial step in solving any trigonometric equation is to isolate the trigonometric function itself. In our equation, $-8 \sqrt{2} \csc (\theta)+6=-10$, the trigonometric function is $\csc (\theta)$. To isolate it, we'll perform a series of algebraic manipulations, much like we would with any other variable. Think of $\csc (\theta)$ as a single entity we want to get by itself on one side of the equation. We start by subtracting 6 from both sides of the equation to move the constant term: $-8 \sqrt{2} \csc (\theta) = -10 - 6$. This simplifies to $-8 \sqrt{2} \csc (\theta) = -16$. Now, we need to get rid of the coefficient $-8 \sqrt{2}$ that's multiplying the $\csc (\theta)$. To do this, we divide both sides of the equation by $-8 \sqrt{2}$: $\csc (\theta) = \frac{-16}{-8 \sqrt{2}}$. Simplifying this fraction, we find that $\csc (\theta) = \frac{2}{\sqrt{2}}$. Often, it's helpful to rationalize the denominator. We can do this by multiplying the numerator and denominator by $\sqrt{2}$: $\csc (\theta) = \frac{2 \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{2 \sqrt{2}}{2}$. This further simplifies to $\csc (\theta) = \sqrt{2}$. So, after these algebraic steps, our equation has been transformed into a much simpler form: $\csc (\theta) = \sqrt{2}$. This is a significant achievement, as it brings us closer to finding the actual values of $\theta$. Remember, every step we take here is designed to isolate the unknown, making it possible to solve for it. We've successfully isolated $\csc (\theta)$, and now we need to understand what this value tells us about $\theta$.

Understanding the Cosecant Function and its Reciprocal

Now that we've successfully isolated the cosecant function and found that $\csc (\theta) = \sqrt{2}$, it's essential to understand the relationship between cosecant and its reciprocal, the sine function. The cosecant function, $\csc (\theta)$, is defined as the reciprocal of the sine function, $\sin (\theta)$. Mathematically, this relationship is expressed as $\csc (\theta) = \frac{1}{\sin (\theta)}$. This identity is a cornerstone of trigonometry and allows us to convert our equation involving cosecant into an equation involving sine, which is often more familiar. Substituting our found value, we get $\sqrt{2} = \frac{1}{\sin (\theta)}$. To solve for $\sin (\theta)$, we can take the reciprocal of both sides of the equation. If $\sqrt{2} = \frac{1}{\sin (\theta)}$, then $\sin (\theta) = \frac{1}{\sqrt{2}}$. Just as before, it's good practice to rationalize the denominator. Multiplying the numerator and denominator by $\sqrt{2}$, we get $\sin (\theta) = \frac{1 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{\sqrt{2}}{2}$. So, our original equation $-8 \sqrt{2} \csc (\theta)+6=-10$ has now been transformed into the much more recognizable trigonometric equation $\sin (\theta) = \frac{\sqrt{2}}{2}$. This transformation is incredibly powerful because the values of sine for common angles are well-known and can be easily recalled from the unit circle or special triangles. The fact that $\sin (\theta)$ is positive tells us that our solutions for $\theta$ must lie in the quadrants where the sine function is positive. On the unit circle, the sine value corresponds to the y-coordinate. The y-coordinate is positive in Quadrant I and Quadrant II. Therefore, we know that our solutions for $\theta$ will be located in these two quadrants. This understanding of reciprocal identities is vital for simplifying and solving a wide array of trigonometric problems, turning complex-looking equations into manageable ones.

Finding Solutions on the Unit Circle

We've arrived at the equation $\sin (\theta) = \frac{\sqrt{2}}{2}$, and we need to find the values of $\theta$ on the interval $[0, 2\pi)$ that satisfy this. The sine function represents the y-coordinate on the unit circle. We are looking for angles where the y-coordinate is $\frac{\sqrt{2}}{2}$. From our knowledge of special right triangles, specifically the 45-45-90 triangle, we know that the sine of 45 degrees (or $\frac{\pi}{4}$ radians) is $\frac{\sqrt{2}}{2}$. This gives us our first solution in Quadrant I: $\theta = \frac{\pi}{4}$. Now, we need to consider the interval $[0, 2\pi)$, which represents one full rotation around the unit circle. Since the sine function is positive in both Quadrant I and Quadrant II, we expect another solution. In Quadrant II, the angle that has the same sine value as $\frac{\pi}{4}$ can be found by subtracting the reference angle from $\pi$ (or 180 degrees). So, the second solution is $\theta = \pi - \frac{\pi}{4}$. To calculate this, we find a common denominator: $\theta = \frac{4\pi}{4} - \frac{\pi}{4} = \frac{3\pi}{4}$. Both of these angles, $\frac{\pi}{4}$ and $\frac{3\pi}{4}$, lie within our specified interval $[0, 2\pi)$. Let's double-check. $\frac{\pi}{4}$ is clearly between 0 and $2\pi$. Similarly, $\frac{3\pi}{4}$ is also between 0 and $2\pi$. Since sine is negative in Quadrants III and IV, there are no other solutions for $\sin (\theta) = \frac{\sqrt{2}}{2}$ within the interval $[0, 2\pi)$. The unit circle is an invaluable tool here, visually representing all possible angle values and their corresponding trigonometric function outputs. By understanding the reference angles and the signs of trigonometric functions in each quadrant, we can accurately determine all solutions within a given interval. The symmetry of the unit circle is key to finding these multiple solutions.

Verifying the Solutions

It's always a good practice, especially in trigonometry, to verify your solutions by plugging them back into the original equation. This helps ensure that you haven't made any algebraic errors or misinterpretations along the way. Our original equation was $-8 \sqrt{2} \csc (\theta)+6=-10$, and we found two potential solutions: $\theta = \frac{\pi}{4}$ and $\theta = \frac{3\pi}{4}$. Let's test the first solution, $\theta = \frac{\pi}{4}$. We know that $\csc (\theta) = \frac{1}{\sin (\theta)}$. So, $\csc (\frac{\pi}{4}) = \frac{1}{\sin (\frac{\pi}{4})} = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$. Now substitute this back into the original equation: $-8 \sqrt{2} (\sqrt{2}) + 6 = -8(2) + 6 = -16 + 6 = -10$. This matches the right side of the equation, so $\theta = \frac{\pi}{4}$ is indeed a valid solution. Now let's test the second solution, $\theta = \frac{3\pi}{4}$. We need to find $\csc (\frac{3\pi}{4})$. We know that $\sin (\frac{3\pi}{4})$ is also $\frac{\sqrt{2}}{2}$ (since $\frac{3\pi}{4}$ is in Quadrant II where sine is positive, and its reference angle is $\frac{\pi}{4}$). Therefore, $\csc (\frac{3\pi}{4}) = \frac{1}{\sin (\frac{3\pi}{4})} = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$. Substituting this into the original equation: $-8 \sqrt{2} (\sqrt{2}) + 6 = -8(2) + 6 = -16 + 6 = -10$. This also matches the right side of the equation, confirming that $\theta = \frac{3\pi}{4}$ is also a valid solution. The verification process is a critical checkpoint. It reinforces your understanding and builds confidence in your answers. If one of the solutions hadn't worked, it would indicate a need to re-examine the steps, looking for errors in algebra or in applying trigonometric identities and unit circle properties. In this case, both solutions hold up perfectly.

Conclusion and Further Exploration

We have successfully navigated the process of solving the trigonometric equation $-8 \sqrt{2} \csc (\theta)+6=-10$ on the interval $[0, 2\pi)$. Through a series of algebraic steps, we isolated the cosecant function, transformed it into its reciprocal sine function, and utilized our knowledge of the unit circle to identify the specific angles that satisfy the equation. Our determined solutions are $\theta = \frac{\pi}{4}$ and $\theta = \frac{3\pi}{4}$. This problem highlights the importance of understanding reciprocal identities and the values of trigonometric functions for common angles. The unit circle serves as an indispensable visual aid, helping us to locate solutions in the correct quadrants and account for all possibilities within the specified interval. Solving trigonometric equations is a skill that is honed with practice, and each problem solved builds a stronger foundation for more complex mathematical endeavors. Remember, the key steps involve isolating the trigonometric function, using identities to simplify, and then employing unit circle knowledge or graphs to find the angles. For those interested in delving deeper into the fascinating world of trigonometry and its applications, exploring resources that cover trigonometric identities, graphs of trigonometric functions, and solving various types of trigonometric equations would be highly beneficial. You can find more in-depth explanations and practice problems on reputable mathematical websites and textbooks. A great resource for further exploration is Khan Academy's section on trigonometry, which offers comprehensive lessons and exercises.