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

Have you ever encountered a situation where a graphing calculator, like Desmos, gives you a solution that seems…off? Specifically, we're diving into why Desmos might show $x = 3\pi/2$ as a solution to the equation $\tan x + \sec x = 2\cos x$, even though the left-hand side of the equation is undefined at that point. This is a fascinating issue that touches on trigonometry, function graphing, and the intricacies of how software handles mathematical equations. Let's break it down, guys!

Understanding the Problem

When we're trying to find solutions to equations like $\tan x + \sec x = 2\cos x$, what we're really looking for are the values of x that make the equation true. The challenge comes when we deal with trigonometric functions that have undefined points. Think about it: $\tan x$ is $\sin x / \cos x$, and $\sec x$ is $1 / \cos x$. So, whenever $\cos x$ is zero, both $\tan x$ and $\sec x$ become undefined. This happens at $x = \pi/2$ and $x = 3\pi/2$, and infinitely many other points as well.

So, why does Desmos sometimes show these undefined points as solutions? Well, it boils down to how these graphing tools work under the hood. Desmos, and similar software, plot points by calculating the function's value at many different x-values and connecting the dots. Sometimes, due to numerical approximations and the way the software handles discontinuities, it might appear that the equation holds true at a point where it technically shouldn't. This is particularly true if the function is very close to zero at a point of discontinuity, leading to potential misinterpretations by the algorithm. It’s kind of like a mirage in the desert – it looks like water, but it’s just an illusion!

This is a crucial point when solving trigonometric equations. We can't just blindly trust the output of a calculator or graphing software. We need to understand the underlying math and be aware of potential pitfalls. This means checking for extraneous solutions, especially when dealing with functions that have denominators or radicals. So, let’s roll up our sleeves and dive into solving the equation ourselves, shall we?

Solving the Equation $\tan x + \sec x = 2\cos x$

Let’s get our hands dirty and solve this equation step by step, so we can truly understand what's going on. Our main goal here is to find all values of x that satisfy the equation $\tan x + \sec x = 2\cos x$ within the interval $[0, 2\pi]$. This will help us understand why Desmos might be giving us that pesky $x = 3\pi/2$ solution.

1. Rewrite in terms of sine and cosine: The first thing we want to do is rewrite everything in terms of sine and cosine, as this will make the equation easier to manipulate. Remember that $\tan x = \frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$. Substituting these into our equation, we get:

   $$\frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2\cos x$$

2. Combine fractions: Now, let's combine the fractions on the left-hand side since they have a common denominator:

   $$\frac{\sin x + 1}{\cos x} = 2\cos x$$

3. Multiply both sides by $\cos x$: To get rid of the fraction, we multiply both sides of the equation by $\cos x$. But, and this is a big but, we need to remember that $\cos x$ cannot be zero. Multiplying gives us:

   $$\sin x + 1 = 2\cos^2 x$$

4. Use the Pythagorean identity: We have $\cos^2 x$ in our equation, and we know the Pythagorean identity $\sin^2 x + \cos^2 x = 1$. We can rearrange this to get $\cos^2 x = 1 - \sin^2 x$. Substituting this into our equation, we get:

   $$\sin x + 1 = 2(1 - \sin^2 x)$$

5. Rearrange into a quadratic: Let's expand and rearrange the equation into a quadratic form. This will allow us to solve for $\sin x$:

   $$\sin x + 1 = 2 - 2\sin^2 x$$

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

6. Solve the quadratic: Now we have a quadratic equation in terms of $\sin x$. We can factor this equation:

   $$(2\sin x - 1)(\sin x + 1) = 0$$

   This gives us two possible solutions for $\sin x$:

   • $2\sin x - 1 = 0 \Rightarrow \sin x = \frac{1}{2}$
   • $\sin x + 1 = 0 \Rightarrow \sin x = -1$

7. Find the values of x: Now we need to find the values of x in the interval $[0, 2\pi]$ that satisfy these two equations:

   • For $\sin x = \frac{1}{2}$, the solutions are $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$. These are the angles in the first and second quadrants where sine is positive and equal to $\frac{1}{2}$.
   • For $\sin x = -1$, the solution is $x = \frac{3\pi}{2}$. This is the angle on the unit circle where the y-coordinate is -1.

8. Check for extraneous solutions: This is the most crucial step! Remember that we multiplied both sides of the equation by $\cos x$ earlier. This means we introduced the possibility of extraneous solutions, values of x that satisfy the transformed equation but not the original one. We need to check our solutions in the original equation $\tan x + \sec x = 2\cos x$.

   • Let's check $x = \frac{\pi}{6}$:

     $$\tan\frac{\pi}{6} + \sec\frac{\pi}{6} = \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$$

     $$2\cos\frac{\pi}{6} = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$$

     So, $x = \frac{\pi}{6}$ is a valid solution.

   • Let's check $x = \frac{5\pi}{6}$:

     $$\tan\frac{5\pi}{6} + \sec\frac{5\pi}{6} = -\frac{1}{\sqrt{3}} - \frac{2}{\sqrt{3}} = -\frac{3}{\sqrt{3}} = -\sqrt{3}$$

     $$2\cos\frac{5\pi}{6} = 2 \cdot \left(-\frac{\sqrt{3}}{2}\right) = -\sqrt{3}$$

     So, $x = \frac{5\pi}{6}$ is also a valid solution.

   • Let's check $x = \frac{3\pi}{2}$: Here's where the problem lies. Both $\tan x$ and $\sec x$ are undefined at $x = \frac{3\pi}{2}$ because $\cos\frac{3\pi}{2} = 0$. Therefore, $x = \frac{3\pi}{2}$ is not a valid solution.

So, after all that work, we find that the valid solutions to the equation $\tan x + \sec x = 2\cos x$ in the interval $[0, 2\pi]$ are $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$. This highlights the importance of checking for extraneous solutions, especially in trigonometric equations!

Why Desmos Gets Confused

Okay, so we've established that $x = 3\pi/2$ isn't a real solution. But that still leaves us with the original question: Why does Desmos show it as one? The answer, as we hinted at earlier, lies in how graphing software handles discontinuities and numerical approximations.

Desmos plots graphs by calculating function values at discrete points and then connecting those points with lines. Near a vertical asymptote (like the one at $x = 3\pi/2$ for $\tan x$ and $\sec x$), the function values shoot off towards infinity (or negative infinity). This creates a steep, almost vertical line on the graph. Desmos might connect points on either side of the asymptote, giving the visual impression that the function exists at $x = 3\pi/2$, even though it technically doesn't.

Moreover, Desmos uses numerical methods to solve equations. These methods involve iterative calculations, and sometimes, due to rounding errors and approximations, the software might converge on a value that is very close to a solution but not exactly a solution. In this case, the value might be close enough to $3\pi/2$ that Desmos flags it as a solution, even though the function is undefined there.

Think of it like this: Desmos is trying its best to paint a picture of the equation, but it's using a limited set of tools and approximations. It's like trying to draw a perfect circle with a slightly wobbly compass – you'll get something that looks like a circle, but it won't be mathematically perfect.

The Takeaway: Critical Thinking in Math

The key takeaway here isn't to bash Desmos or any other graphing tool. These tools are incredibly powerful and useful for visualizing and exploring mathematical concepts. However, they are just tools, and like any tool, they have limitations. We, as math students and enthusiasts, need to be critical thinkers. We can't just blindly accept the output of a calculator or software without understanding the underlying mathematics.

This example with $\tan x + \sec x = 2\cos x$ is a fantastic illustration of this principle. It shows us that:

• Undefined points matter: We need to be aware of the domains of functions and where they are undefined.
• Extraneous solutions are a thing: When we manipulate equations, especially by multiplying or dividing by expressions that could be zero, we need to check our solutions.
• Software has limitations: Graphing tools and calculators use numerical methods and approximations, which can lead to errors.
• Understanding the math is crucial: The best way to avoid these pitfalls is to have a solid understanding of the underlying mathematical concepts.

So, the next time you're solving a trigonometric equation (or any equation, for that matter), remember this example. Use Desmos and other tools to help you visualize and explore, but always think critically and verify your solutions. Happy math-ing, guys!

Main Question Revisited

The main question posed was: how many values of x satisfy the equation $\tan x + \sec x = 2\cos x$ in the interval $[0, 2\pi]$? As we've shown through our step-by-step solution, there are indeed only two valid solutions: $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$. The value $x = \frac{3\pi}{2}$, while it might appear as a solution on Desmos due to the software's approximations, is actually an extraneous solution because the original equation is undefined at that point.

This underscores the importance of not only solving the equation algebraically but also verifying the solutions by plugging them back into the original equation and checking for any inconsistencies or undefined terms. Remember, mathematical rigor is key to ensuring accurate results!