Solve 2 Sec^2 X - Tan^4 X = 3

by Andrew McMorgan 30 views

Hey math whizzes and fellow equation explorers! Today, we're diving deep into the world of trigonometry to tackle a problem that looks a bit intimidating at first glance but is totally conquerable with the right approach. We're going to unravel the exact solutions for the equation $2 \sec^2 x - \tan^4 x = 3$, and guess what? We need our answers in radians. So, grab your calculators, your favorite study snacks, and let's get this done!

Understanding the Equation: The Basics

Alright guys, let's break down the equation: $2 \sec^2 x - \tan^4 x = 3$. Our mission, should we choose to accept it, is to find all the values of $x$ (in radians, remember!) that make this equation true. This involves playing around with trigonometric identities to simplify the equation into a form we can solve more easily. The key players here are the secant ($\sec x$) and tangent ($\tan x$) functions. We know that $\sec x = \frac{1}{\cos x}$ and $\tan x = \frac{\sin x}{\cos x}$. We also have a super handy Pythagorean identity that relates these guys: $1 + \tan^2 x = \sec^2 x$. This identity is going to be our best friend in simplifying this beast of an equation. It allows us to express everything in terms of a single trigonometric function, which makes solving so much smoother. So, before we even touch the given equation, let's get cozy with this identity because it's the golden ticket to unlocking the solutions.

The Substitution Strategy: Making it Simpler

Now, let's get strategic. Our equation has both $\sec^2 x$ and $\tan^4 x$. To make things less confusing, we want to express everything using just one of these functions. Since we have the identity $1 + \tan^2 x = \sec^2 x$, it's easiest to substitute $\sec^2 x$ with $1 + \tan^2 x$. Let's do that!

Our original equation is:

2secā”2xāˆ’tanā”4x=32 \sec^2 x - \tan^4 x = 3

Substitute $ \sec^2 x $ with $ 1 + \tan^2 x $:

2(1+tanā”2x)āˆ’tanā”4x=32 (1 + \tan^2 x) - \tan^4 x = 3

Now, let's expand and rearrange this to see what we get:

2+2tanā”2xāˆ’tanā”4x=32 + 2 \tan^2 x - \tan^4 x = 3

We want to get everything on one side to form a polynomial-like equation in terms of $\tan x$. Let's move all terms to the right side to make the $\tan^4 x$ term positive:

0=tanā”4xāˆ’2tanā”2x+3āˆ’20 = \tan^4 x - 2 \tan^2 x + 3 - 2

0=tanā”4xāˆ’2tanā”2x+10 = \tan^4 x - 2 \tan^2 x + 1

Look at that! This looks way more manageable. If we let $y = \tan^2 x$, the equation transforms into:

y2āˆ’2y+1=0y^2 - 2y + 1 = 0

This is a perfect quadratic equation, and it's actually a perfect square trinomial! It factors beautifully.

Factoring and Solving for $\tan^2 x$

So, we have $y^2 - 2y + 1 = 0$. This factors as:

(yāˆ’1)2=0(y - 1)^2 = 0

This means $y - 1 = 0$, which gives us $y = 1$.

Remember that we set $y = \tan^2 x$? So, we now have:

tanā”2x=1\tan^2 x = 1

This is a significant step! We've simplified our complex trigonometric equation into a much simpler one involving the tangent function. This step is crucial because it isolates the core trigonometric relationship we need to solve. It's like getting to the heart of the matter. When you encounter complex equations like the original one, always look for opportunities to use identities to simplify them. This often involves substituting one trigonometric function for another, aiming to reduce the number of different functions or powers present. In this case, we successfully converted an equation involving secant and powers of tangent into a simple equation involving the square of the tangent.

Finding the Values of $\tan x$

From $\tan^2 x = 1$, we can now find the possible values for $\tan x$. Taking the square root of both sides, we get:

tanā”x=Ā±1\tan x = \pm \sqrt{1}

tanā”x=Ā±1\tan x = \pm 1

This means we have two separate cases to consider:

  1. tanā”x=1\tan x = 1

  2. tanā”x=āˆ’1\tan x = -1

These two conditions will lead us to all the solutions for our original equation. It's important to remember that when you take the square root of both sides of an equation like $a^2 = b$, you must consider both the positive and negative roots, $a = \sqrt{b}$ and $a = -\sqrt{b}$. This is precisely why we split our problem into two cases here, ensuring we don't miss any potential solutions.

Case 1: $\tan x = 1$

When $\tan x = 1$, we need to find the angles $x$ (in radians) whose tangent is 1. We know that the tangent function represents the ratio $\frac{\sin x}{\cos x}$. Tangent is positive in the first and third quadrants. The principal value (the angle in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$) for which $\tan x = 1$ is $x = \frac{\pi}{4}$.

Since the tangent function has a period of $\pi$, meaning $\tan(x + k\pi) = \tan x$ for any integer $k$, the general solution for $\tan x = 1$ is:

x=Ļ€4+kĻ€x = \frac{\pi}{4} + k \pi

where $k$ is any integer.

Let's visualize this on the unit circle. Tangent is positive in Quadrant I and Quadrant III. The angle in Quadrant I is $\frac{\pi}{4}$. To get to the angle in Quadrant III that has the same tangent value, we add $\pi$ to $\frac{\pi}{4}$, which gives us $\frac{5\pi}{4}$. If we keep adding or subtracting multiples of $\pi$, we cover all angles where the tangent is 1. This is why the $+ k\pi$ covers all solutions.

Case 2: $\tan x = -1$

Next, we tackle the case where $\tan x = -1$. Tangent is negative in the second and fourth quadrants. The principal value for which $\tan x = -1$ is $x = -\frac{\pi}{4}$.

Alternatively, we can think of the reference angle as $\frac{\pi}{4}$. Since tangent is negative in Quadrant II and Quadrant IV, the angles are $\pi - \frac{\pi}{4} = \frac{3\pi}{4}$ and $2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$.

Using the periodicity of the tangent function (period is $\pi$), the general solution for $\tan x = -1$ is:

x=āˆ’Ļ€4+kĻ€x = -\frac{\pi}{4} + k \pi

where $k$ is any integer.

We can also write this as $x = \frac{3\pi}{4} + k \pi$, because $- \frac{\pi}{4} + \pi = \frac{3\pi}{4}$. Both forms are correct and represent the same set of solutions.

On the unit circle, $\tan x = -1$ occurs in Quadrant II (at $\frac{3\pi}{4}$) and Quadrant IV (at $\frac{7\pi}{4}$, which is equivalent to $-\frac{\pi}{4}$ plus $2\pi$). Adding or subtracting multiples of $\pi$ from these angles will generate all possible solutions for this case. For example, $\frac{3\pi}{4} + \pi = \frac{7\pi}{4}$, and $\frac{7\pi}{4} + \pi = \frac{11\pi}{4}$, which is also an angle where tangent is -1.

Combining All Solutions

So, we have two sets of general solutions:

  • From $\tan x = 1$: $x = \frac{\pi}{4} + k \pi$
  • From $\tan x = -1$: $x = -\frac{\pi}{4} + k \pi$ (or $x = \frac{3\pi}{4} + k \pi$)

Now, let's look at the options provided in the original question (which seem to be missing from your prompt, but I'll assume they are standard multiple-choice options typically seen in such problems). The goal is to see which option neatly encapsulates all these solutions.

Let's analyze the structure of our solutions. We have angles spaced $\pi$ apart where the tangent is 1, and angles spaced $\pi$ apart where the tangent is -1. It's important to note that these sets of solutions are distinct. The values of $x$ for $\tan x = 1$ are $\frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, ...$ and $-\frac{3\pi}{4}, -\frac{7\pi}{4}, ...$. The values of $x$ for $\tan x = -1$ are $-\frac{\pi}{4}, \frac{3\pi}{4}, \frac{7\pi}{4}, ...$ and $-\frac{5\pi}{4}, -\frac{9\pi}{4}, ...$.

If we consider the angles within one full circle ($0$ to $2\pi$), the solutions for $\tan x = 1$ are $\frac{\pi}{4}$ and $\frac{5\pi}{4}$. The solutions for $\tan x = -1$ are $\frac{3\pi}{4}$ and $\frac{7\pi}{4}$.

Let's re-examine the provided options (assuming they are standard forms):

A. $\frac{\pi}{3}+k \pi$ and $\frac{2 \pi}{3}+k \pi$ B. $\frac{\pi}{3}+2 k \pi$ and $\frac{5 \pi}{3}+2 k \pi$

Neither of these options directly match our derived solutions of $x = \frac{\pi}{4} + k \pi$ and $x = \frac{3\pi}{4} + k \pi$. It seems there might be a misunderstanding or a typo in the provided options, as our derived solutions are based on solid trigonometric principles. Let's double check our work.

Verification Step: Are there other identities or interpretations?

We used $1 + \tan^2 x = \sec^2 x$. What if we had expressed everything in terms of $\cos x$?

2secā”2xāˆ’tanā”4x=32 \sec^2 x - \tan^4 x = 3

2(1cosā”2x)āˆ’(sinā”2xcosā”2x)2=32 \left(\frac{1}{\cos^2 x}\right) - \left(\frac{\sin^2 x}{\cos^2 x}\right)^2 = 3

2cosā”2xāˆ’sinā”4xcosā”4x=3 \frac{2}{\cos^2 x} - \frac{\sin^4 x}{\cos^4 x} = 3

Multiply by $ \cos^4 x $:

2cosā”2xāˆ’sinā”4x=3cosā”4x2 \cos^2 x - \sin^4 x = 3 \cos^4 x

Replace $\sin^2 x$ with $1-\cos^2 x$:

2cosā”2xāˆ’(1āˆ’cosā”2x)2=3cosā”4x2 \cos^2 x - (1-\cos^2 x)^2 = 3 \cos^4 x

2cosā”2xāˆ’(1āˆ’2cosā”2x+cosā”4x)=3cosā”4x2 \cos^2 x - (1 - 2 \cos^2 x + \cos^4 x) = 3 \cos^4 x

2cosā”2xāˆ’1+2cosā”2xāˆ’cosā”4x=3cosā”4x2 \cos^2 x - 1 + 2 \cos^2 x - \cos^4 x = 3 \cos^4 x

4cosā”2xāˆ’1āˆ’cosā”4x=3cosā”4x4 \cos^2 x - 1 - \cos^4 x = 3 \cos^4 x

4cosā”4xāˆ’4cosā”2x+1=04 \cos^4 x - 4 \cos^2 x + 1 = 0

Let $z = \cos^2 x$. Then $4z^2 - 4z + 1 = 0$. This is $(2z - 1)^2 = 0$. So, $2z - 1 = 0$, which means $z = \frac{1}{2}$. Substituting back $z = \cos^2 x$:

cosā”2x=12\cos^2 x = \frac{1}{2}

cosā”x=Ā±12=Ā±22\cos x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}

Now, let's find the solutions for $\cos x = \frac{\sqrt{2}}{2}$ and $\cos x = -\frac{\sqrt{2}}{2}$ in radians.

For $\cos x = \frac{\sqrt{2}}{2}$, the principal value is $x = \frac{\pi}{4}$. Since cosine is positive in Quadrants I and IV, the solutions in $[0, 2\pi)$ are $x = \frac{\pi}{4}$ and $x = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$. The general solution is $x = \pm \frac{\pi}{4} + 2k\pi$.

For $\cos x = -\frac{\sqrt{2}}{2}$, the principal value is $x = \frac{3\pi}{4}$. Since cosine is negative in Quadrants II and III, the solutions in $[0, 2\pi)$ are $x = \frac{3\pi}{4}$ and $x = 2\pi - \frac{3\pi}{4} = \frac{5\pi}{4}$. The general solution is $x = \pm \frac{3\pi}{4} + 2k\pi$.

Let's combine these. The set of all solutions is:

x=Ļ€4+2kĻ€x = \frac{\pi}{4} + 2k\pi

x=7Ļ€4+2kĻ€x = \frac{7\pi}{4} + 2k\pi

x=3Ļ€4+2kĻ€x = \frac{3\pi}{4} + 2k\pi

x=5Ļ€4+2kĻ€x = \frac{5\pi}{4} + 2k\pi

Notice that $\frac{7\pi}{4}$ is equivalent to $-\frac{\pi}{4}$, so $x = \pm \frac{\pi}{4} + 2k\pi$. And $\frac{5\pi}{4}$ is equivalent to $-\frac{3\pi}{4}$, so $x = \pm \frac{3\pi}{4} + 2k\pi$.

Now, let's compare these general solutions with the tangent solutions we found earlier:

  • From $\tan x = 1$: $x = \frac{\pi}{4} + k \pi$. This covers $ \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, ... $
  • From $\tan x = -1$: $x = \frac{3\pi}{4} + k \pi$. This covers $ \frac{3\pi}{4}, \frac{7\pi}{4}, \frac{11\pi}{4}, ... $

These two sets ($\frac{\pi}{4} + k \pi$ and $\frac{3\pi}{4} + k \pi$) indeed encompass all the solutions derived from the cosine equation. For example, $ \frac{\pi}{4} + k \pi $ gives $\frac{\pi}{4}$ (for $k=0$) and $\frac{5\pi}{4}$ (for $k=1$). And $ \frac{3\pi}{4} + k \pi $ gives $\frac{3\pi}{4}$ (for $k=0$) and $\frac{7\pi}{4}$ (for $k=1$).

Therefore, the general solutions are $x = \frac{\pi}{4} + k \pi$ and $x = \frac{3\pi}{4} + k \pi$.

Now, let's reconsider the options given in the problem statement. The question asks which shows ALL the exact solutions. The options provided were:

A. $\frac{\pi}{3}+k \pi$ and $\frac{2 \pi}{3}+k \pi$ B. $\frac{\pi}{3}+2 k \pi$ and $\frac{5 \pi}{3}+2 k \pi$

It appears there is a significant mismatch between the derived correct solutions ($\frac{\pi}{4}+k \pi$ and $\frac{3\pi}{4}+k \pi$) and the provided options. This suggests either the options are incorrect, or there's a misunderstanding of the question's intent. However, based on the standard mathematical interpretation and derivation, the solutions we found are correct.

Let's assume there was a typo in the question and the equation was different, or the options were meant to be different. If we HAVE to choose from the given options, it implies that our derived solutions must match one of them. Since they don't, let's re-evaluate if there's any scenario where options A or B could arise.

Option A has $k\pi$ periodicity. This means the tangent function is likely involved. The angles $\frac{\pi}{3}$ and $\frac{2\pi}{3}$ have tangent values of $\sqrt{3}$ and $-\sqrt{3}$ respectively. If $\tan x = \pm \sqrt{3}$, then $\tan^2 x = 3$. Our equation led to $\tan^2 x = 1$. So, Option A is incorrect for the given equation.

Option B has $2k\pi$ periodicity. This is typical for sine and cosine equations where we consider solutions within $0$ to $2\pi$ (or $- \pi$ to $\pi$) and then add multiples of $2\pi$. The angles $\frac{\pi}{3}$ and $\frac{5\pi}{3}$ (which is $-\frac{\pi}{3}$) are related to $\cos x = \frac{1}{2}$. If $\cos x = \pm \frac{1}{2}$, then $\cos^2 x = \frac{1}{4}$. Our derived equation from cosine was $\cos^2 x = \frac{1}{2}$. So, Option B is also incorrect for the given equation.

Given the strong derivation that $\tan^2 x = 1$ and consequently $\cos^2 x = 1/2$, leading to solutions $x = \frac{\pi}{4} + k \pi$ and $x = \frac{3\pi}{4} + k \pi$, it is highly probable that the provided options A and B are incorrect for the equation $2 \sec^2 x - \tan^4 x = 3$.

Final Conclusion: Based on rigorous mathematical derivation using trigonometric identities, the exact solutions in radians for the equation $2 \sec^2 x - \tan^4 x = 3$ are given by $x = \frac{\pi}{4} + k \pi$ and $x = \frac{3\pi}{4} + k \pi$, where $k$ is any integer. The options provided (A and B) do not align with these correct solutions. If forced to select from the given options, it would indicate an error in the question or the options themselves.