Desmos' False Solution: Tan X + Sec X = 2cos X Explained

Introduction

Hey guys! Let's dive into a fascinating puzzle involving trigonometry, graphing functions, and the ever-helpful Desmos graphing calculator. We're tackling the equation $\tan x + \sec x = 2\cos x$, and something quirky happens when we look for solutions. Specifically, we want to understand why Desmos indicates that $x = \frac{3\pi}{2}$ is a solution, even though the left-hand side of the equation isn't even defined at that point. It's like Desmos is trying to pull a fast one on us, but let's get to the bottom of this trigonometric mystery. We'll explore the nuances of trigonometric functions, their domains, and how graphing calculators handle such situations. This is crucial for anyone learning trigonometry, graphing functions, or using tools like Desmos, as it highlights the importance of understanding the underlying math, not just blindly trusting the technology. So, buckle up, and let's unravel this mathematical conundrum together!

Understanding the Problem

When dealing with trigonometric equations, it's essential to remember that not all expressions are created equal, especially when it comes to their domains. The domain of a function is the set of all possible input values (in this case, values of x) for which the function is defined. Our equation involves $\tan x$ and $\sec x$, both of which have restrictions on their domains. Specifically, $\tan x = \frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$. This means that whenever $\cos x = 0$, both $\tan x$ and $\sec x$ are undefined. This is a critical point to keep in mind as we analyze the equation. Why? Because at these undefined points, the equation simply doesn't make sense. It's like trying to divide by zero – it breaks the fundamental rules of mathematics. The value $x = \frac{3\pi}{2}$ is a classic example. At this point, $\cos(\frac{3\pi}{2}) = 0$, making both $\tan(\frac{3\pi}{2})$ and $\sec(\frac{3\pi}{2})$ undefined. So, if these terms are undefined, how can Desmos possibly show this as a solution? That's the core of our puzzle, and we're going to dissect it piece by piece. This exploration isn't just an academic exercise; it's about developing a deep, intuitive understanding of trigonometry and the tools we use to explore it. By understanding the limitations and quirks of graphing calculators like Desmos, we become better problem-solvers and more critical thinkers in mathematics.

Trigonometric Functions and Their Domains

Let's dive a bit deeper into trigonometric functions and their domains. As we touched on earlier, the functions $\tan x$ and $\sec x$ are defined in terms of $\sin x$ and $\cos x$. Specifically, $\tan x = \frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$. The cosine function, $\cos x$, oscillates between -1 and 1, and it equals zero at certain points. These points are where our trouble begins. Think of the unit circle – $\cos x$ represents the x-coordinate of a point on the circle. $\cos x$ is zero at angles where the point lies on the vertical axis, which occurs at $x = \frac{\pi}{2} + n\pi$, where n is an integer. This means $\cos x$ is zero at $\frac{\pi}{2}$, $\frac{3\pi}{2}$, $\frac{5\pi}{2}$, and so on, as well as at negative counterparts like $-\frac{\pi}{2}$, $-\frac{3\pi}{2}$, etc. Now, because $\tan x$ and $\sec x$ both have $\cos x$ in the denominator, they are undefined at these same points. This is a crucial concept: whenever we have a fraction, we must ensure the denominator is not zero. In our equation, $\tan x + \sec x = 2\cos x$, the left-hand side is undefined at $x = \frac{3\pi}{2}$ because both $\tan(\frac{3\pi}{2})$ and $\sec(\frac{3\pi}{2})$ are undefined. This is a non-negotiable mathematical rule. So, even before we start solving the equation, we know that $x = \frac{3\pi}{2}$ cannot be a valid solution. Understanding these domain restrictions is paramount. It prevents us from making fundamental errors and ensures that we interpret the results from tools like Desmos correctly. It’s a cornerstone of trigonometric problem-solving, reminding us that mathematical rigor is just as important as the computational power we wield.

Solving the Equation

Okay, now let's get our hands dirty and solve the equation $\tan x + \sec x = 2\cos x$. This is where things get interesting, and we'll see why Desmos might lead us astray if we're not careful. To solve this, we need to manipulate the equation using trigonometric identities. Recall that $\tan x = \frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$. Substituting these into the equation, we get: $\frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2\cos x$. Now, we can combine the fractions on the left side: $\frac{\sin x + 1}{\cos x} = 2\cos x$. To get rid of the fraction, we can multiply both sides by $\cos x$, but here's a critical caveat: we are assuming $\cos x \neq 0$. We already know why – because if $\cos x = 0$, the original equation is undefined. Multiplying, we get: $\sin x + 1 = 2\cos^2 x$. Now, we need to express everything in terms of a single trigonometric function. We can use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ to rewrite $\cos^2 x$ as $1 - \sin^2 x$. Substituting, we have: $\sin x + 1 = 2(1 - \sin^2 x)$. Expanding and rearranging, we get a quadratic equation in terms of $\sin x$: $2\sin^2 x + \sin x - 1 = 0$. This quadratic can be factored: $(2\sin x - 1)(\sin x + 1) = 0$. Setting each factor equal to zero gives us two possible solutions: $2\sin x - 1 = 0 \implies \sin x = \frac{1}{2}$ and $\sin x + 1 = 0 \implies \sin x = -1$. Now, we need to find the values of x in the interval $[0, 2\pi]$ that satisfy these equations. For $\sin x = \frac{1}{2}$, we have two solutions: $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$. For $\sin x = -1$, we have one solution: $x = \frac{3\pi}{2}$. But wait! Remember our initial caveat? We assumed $\cos x \neq 0$. Since $\cos(\frac{3\pi}{2}) = 0$, this solution is extraneous. It arises from our algebraic manipulation but doesn't satisfy the original equation's domain. So, the valid solutions are $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$. This careful approach to solving the equation highlights the importance of considering domain restrictions at every step. It's not just about finding numbers that satisfy a transformed equation; it's about finding solutions that make sense in the original context.

Why Desmos Shows the "Solution"

So, let's tackle the million-dollar question: Why does Desmos show $x = \frac{3\pi}{2}$ as a solution when it's clearly not? This is where we need to understand how graphing calculators, like Desmos, work their magic – and where they can sometimes lead us astray if we're not careful. Desmos graphs equations by plotting points. It essentially plugs in a bunch of x-values, calculates the corresponding y-values for each side of the equation, and then plots those points. If the y-values are the same (or very, very close), Desmos considers that x-value a solution and marks it on the graph. However, Desmos doesn't always "think" about domain restrictions the way we do as mathematicians. It's a powerful tool, but it's not infallible. In the case of our equation, $\tan x + \sec x = 2\cos x$, Desmos plots both $y = \tan x + \sec x$ and $y = 2\cos x$. Near $x = \frac{3\pi}{2}$, both functions have vertical asymptotes. A vertical asymptote is a vertical line that the graph of a function approaches but never actually touches. Desmos, in its point-plotting approach, might connect points on either side of the asymptote, creating the illusion of an intersection at $x = \frac{3\pi}{2}$. Essentially, Desmos is showing us where the graphs appear to intersect, but it's not accounting for the fact that $\tan x$ and $\sec x$ are undefined at that point. This is a crucial lesson: Graphing calculators are fantastic tools for visualization, but they don't replace mathematical understanding. We need to be able to interpret what we see on the screen and reconcile it with what we know about the functions themselves. Desmos gives us a visual representation, but it's up to us to apply our mathematical knowledge to determine which solutions are valid and which are extraneous. It's a partnership between human reasoning and computational power.

The Importance of Checking for Extraneous Solutions

This whole situation underscores the importance of checking for extraneous solutions in trigonometric (and other) equations. Extraneous solutions are values that pop up during the solving process but don't actually satisfy the original equation. They're like imposters – they look like solutions, but they're not the real deal. In our case, $x = \frac{3\pi}{2}$ is the perfect example of an extraneous solution. It emerged when we multiplied both sides of the equation by $\cos x$, a perfectly valid algebraic step as long as $\cos x \neq 0$. But because $\cos(\frac{3\pi}{2}) = 0$, this step introduced a potential solution that doesn't work in the original equation. So, how do we guard against these sneaky imposters? The key is to always, always check your solutions in the original equation. After you've solved an equation, plug your solutions back into the original equation and see if they make it true. If a solution makes the equation undefined (like in our case) or leads to a contradiction, it's an extraneous solution and must be discarded. Checking for extraneous solutions is not just a formality; it's a fundamental part of the problem-solving process. It's the mathematical equivalent of double-checking your work, ensuring that your answer is not only correct but also meaningful in the context of the problem. This meticulous approach is what separates a good mathematician from a great one. We're not just looking for numbers; we're looking for solutions that make sense.

Main Question: Solutions in the Interval $[0, 2π]$

Let's circle back to the main question: how many values of x satisfy the equation $\tan x + \sec x = 2\cos x$ in the interval $[0, 2\pi]$? We've already done the heavy lifting in solving the equation, so now it's just a matter of summarizing our findings. We found three potential solutions: $x = \frac{\pi}{6}$, $x = \frac{5\pi}{6}$, and $x = \frac{3\pi}{2}$. However, we also discovered that $x = \frac{3\pi}{2}$ is an extraneous solution because it makes the left-hand side of the original equation undefined. This is a critical point! We can't just count the number of potential solutions; we have to weed out the imposters. That leaves us with two valid solutions: $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$. Both of these values lie within the interval $[0, 2\pi]$, and they both satisfy the original equation when we plug them in. So, the final answer is that there are two solutions to the equation $\tan x + \sec x = 2\cos x$ in the interval $[0, 2\pi]$. This question highlights the importance of not only solving equations but also understanding the context and limitations of the solutions. It's a reminder that mathematical problem-solving is not just about finding numbers; it's about critical thinking, logical reasoning, and a deep understanding of the underlying concepts. We started with a quirky observation from Desmos, and we've arrived at a solid, mathematically sound answer. That's the power of combining technology with a strong foundation in mathematics.

Conclusion

Alright, guys, we've journeyed through a fascinating trigonometric puzzle! We started with a seemingly simple equation, $\tan x + \sec x = 2\cos x$, and encountered a perplexing issue: Desmos showed $x = \frac{3\pi}{2}$ as a solution, even though it's not. By diving deep into the domains of trigonometric functions, carefully solving the equation, and understanding how graphing calculators work, we unraveled the mystery. We discovered that $x = \frac{3\pi}{2}$ is an extraneous solution, arising from a point where $\tan x$ and $\sec x$ are undefined. We learned a crucial lesson about the importance of checking for extraneous solutions and not blindly trusting technology. Desmos is a powerful tool, but it's our mathematical understanding that allows us to interpret its output correctly. Ultimately, we found that the equation has two valid solutions in the interval $[0, 2\pi]$: $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$. This exploration is a fantastic example of how mathematical thinking works. It's not just about memorizing formulas and procedures; it's about critical analysis, problem-solving, and a deep appreciation for the nuances of mathematics. So, the next time you're using a graphing calculator or tackling a tricky equation, remember the lessons we've learned here. Trust your mathematical instincts, check your solutions, and never be afraid to ask "why?" That's where the real mathematical understanding begins. Keep exploring, keep questioning, and keep those mathematical gears turning!