Solve (12 Sin^2 X - 4 Cos^2 X) / Sin 2x = 1: Step-by-Step

Hey guys! Today, we're diving into the exciting world of trigonometry to tackle a problem that might seem a bit daunting at first glance. But don't worry, we'll break it down step by step, making it super easy to understand. We're going to solve the trigonometric equation: $\frac{12 \sin ^2 x-4 \cos ^2 x}{\sin 2 x}=1$. So, grab your thinking caps, and let's get started!

Understanding the Problem

Before we jump into the solution, let's make sure we understand what the problem is asking. We have a trigonometric equation involving sine and cosine functions, and our goal is to find the values of x that satisfy this equation. Essentially, we're looking for the angles x where the given expression equals 1. Trigonometric equations pop up in various fields, from physics to engineering, so mastering them is super useful.

When you first look at $\frac{12 \sin ^2 x-4 \cos ^2 x}{\sin 2 x}=1$, it might seem complex. There are squares of sine and cosine, and then there's that $\sin 2x$ in the denominator. But the key to solving any math problem is to break it down into smaller, manageable parts. We'll use trigonometric identities and algebraic manipulation to simplify the equation. Remember, trigonometric identities are our best friends here – they allow us to rewrite trigonometric expressions in different forms, often making them easier to work with. For example, we know that $\sin 2x$ can be expressed in terms of $\sin x$ and $\cos x$, which is a crucial step in solving this problem. So, let's keep these identities in mind as we move forward.

Another important thing to consider is the domain of our equation. Since we have $\sin 2x$ in the denominator, we need to make sure that $\sin 2x$ is not equal to zero. Why? Because division by zero is undefined in mathematics. So, we'll need to keep track of the values of x that would make $\sin 2x = 0$ and exclude them from our final solution. This is a common step in solving trigonometric equations, especially when there are fractions involved. So, before we even start manipulating the equation, we have a little side quest to find these excluded values. This might seem like a small detail, but it's super important for ensuring the accuracy of our solution.

Step-by-Step Solution

1. Simplify the Equation

Our first step is to get rid of the fraction. We can do this by multiplying both sides of the equation by $\sin 2x$. This gives us:

$12 \sin ^2 x - 4 \cos ^2 x = \sin 2x$

Now, we need to express everything in terms of $\sin x$ and $\cos x$. Remember the double angle identity for sine? $\sin 2x = 2 \sin x \cos x$. Let's substitute that in:

$12 \sin ^2 x - 4 \cos ^2 x = 2 \sin x \cos x$

Okay, we're making progress! Everything is now in terms of $\sin x$ and $\cos x$. This is a good sign because we can now start thinking about how to further simplify the equation. One strategy we can use is to try and get all the terms on one side, making the equation equal to zero. This is a common technique in solving equations because it allows us to use factoring or other methods to find the solutions. So, let's move that $2 \sin x \cos x$ term to the left side.

2. Rearrange and Use Trigonometric Identities

Let's move all the terms to one side:

$12 \sin ^2 x - 4 \cos ^2 x - 2 \sin x \cos x = 0$

Now, this looks a bit more manageable. We have a quadratic-like equation in terms of $\sin x$ and $\cos x$. But we still have both $\sin^2 x$ and $\cos^2 x$, which makes it a bit tricky to work with. It would be great if we could express everything in terms of just one trigonometric function. And guess what? We can! Remember the Pythagorean identity? $\sin^2 x + \cos^2 x = 1$. We can use this to express $\cos^2 x$ in terms of $\sin^2 x$ (or vice versa).

Let's rewrite $\cos^2 x$ as $1 - \sin^2 x$:

$12 \sin ^2 x - 4 (1 - \sin ^2 x) - 2 \sin x \cos x = 0$

Now, let's distribute that -4:

$12 \sin ^2 x - 4 + 4 \sin ^2 x - 2 \sin x \cos x = 0$

Combine the $\sin^2 x$ terms:

$16 \sin ^2 x - 2 \sin x \cos x - 4 = 0$

3. Factor and Solve

Now we have a simplified equation. To make it even simpler, let's divide the entire equation by 2:

$8 \sin ^2 x - \sin x \cos x - 2 = 0$

This looks a bit like a quadratic equation, but we have that pesky $\sin x \cos x$ term. To deal with this, let's try to factor the equation or use another clever trick. Sometimes, recognizing patterns or using substitutions can help. In this case, we might try to express everything in terms of tangent, since $\tan x = \frac{\sin x}{\cos x}$. To do this, we can divide the entire equation by $\cos^2 x$, but we need to be careful about the cases where $\cos x = 0$ (since we can't divide by zero).

Let's assume $\cos x \neq 0$ for now and divide by $\cos^2 x$:

$\frac{8 \sin ^2 x}{\cos ^2 x} - \frac{\sin x \cos x}{\cos ^2 x} - \frac{2}{\cos ^2 x} = 0$

This simplifies to:

$8 \tan ^2 x - \tan x - 2 \sec ^2 x = 0$

Remember that $\sec^2 x = 1 + \tan^2 x$, so we can substitute that in:

$8 \tan ^2 x - \tan x - 2 (1 + \tan ^2 x) = 0$

Distribute the -2:

$8 \tan ^2 x - \tan x - 2 - 2 \tan ^2 x = 0$

Combine the $\tan^2 x$ terms:

$6 \tan ^2 x - \tan x - 2 = 0$

Now we have a quadratic equation in terms of $\tan x$! Let's substitute $y = \tan x$ to make it look even more familiar:

$6y^2 - y - 2 = 0$

We can solve this quadratic equation using the quadratic formula, factoring, or any other method you prefer. Let's try factoring:

$(2y + 1)(3y - 2) = 0$

This gives us two possible solutions for y:

$y = -\frac{1}{2}$ or $y = \frac{2}{3}$

Now, remember that $y = \tan x$, so we have:

$\tan x = -\frac{1}{2}$ or $\tan x = \frac{2}{3}$

4. Find the Values of x

Now we need to find the values of x that satisfy these equations. We'll use the inverse tangent function (arctan) to do this.

For $\tan x = -\frac{1}{2}$:

$x = \arctan(-\frac{1}{2}) + n\pi$, where n is an integer.

Using a calculator, we find that $\arctan(-\frac{1}{2}) \approx -0.4636$ radians. So, the general solution is:

$x \approx -0.4636 + n\pi$

For $\tan x = \frac{2}{3}$:

$x = \arctan(\frac{2}{3}) + n\pi$, where n is an integer.

Using a calculator, we find that $\arctan(\frac{2}{3}) \approx 0.5880$ radians. So, the general solution is:

$x \approx 0.5880 + n\pi$

5. Check for Extraneous Solutions

Remember that we assumed $\cos x \neq 0$ when we divided by $\cos^2 x$. We also need to check for values of x that make $\sin 2x = 0$, since that was in the denominator of our original equation. $\sin 2x = 0$ when $2x = n\pi$, or $x = \frac{n\pi}{2}$, where n is an integer. These values are $0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$, and so on.

We need to make sure that our solutions don't include these values. Let's check a few values of n in our general solutions and see if they coincide with $x = \frac{n\pi}{2}$.

Our general solutions are $x \approx -0.4636 + n\pi$ and $x \approx 0.5880 + n\pi$. By plugging in different integer values for n, we can see that none of these solutions coincide with $x = \frac{n\pi}{2}$. This means we don't have any extraneous solutions to worry about.

Final Solution

So, the solutions to the equation $\frac{12 \sin ^2 x-4 \cos ^2 x}{\sin 2 x}=1$ are:

$x \approx -0.4636 + n\pi$ and $x \approx 0.5880 + n\pi$, where n is an integer.

Conclusion

And there you have it! We've successfully solved a trigonometric equation by breaking it down into smaller steps, using trigonometric identities, and carefully checking for extraneous solutions. Solving these kinds of problems is all about practice, so keep at it! Remember, trig identities are your friends, and a systematic approach can make even the trickiest equations manageable. Keep practicing, and you'll become a trig master in no time!