Solving $-3 Tan^2 X + 1 = 0$ Exact Solutions In Radians

Hey everyone! Let's dive into solving a trigonometric equation step-by-step. We've got the equation $-3 \tan^2 x + 1 = 0$, and our mission is to find all the exact solutions in radians. Trigonometric equations might seem daunting at first, but trust me, with a systematic approach, they become quite manageable. We'll break down each step, ensuring you grasp the underlying concepts. So, buckle up and let's get started!

Step 1: Isolate the Trigonometric Function

In this initial step, our main goal is to isolate the trigonometric function, which in this case is $\tan^2 x$. Think of it like solving a regular algebraic equation; we want to get the term with the variable on one side of the equation. So, how do we do that? We start by getting rid of the constant term. We have the equation $-3 \tan^2 x + 1 = 0$. To isolate the term with $\tan^2 x$, we need to subtract 1 from both sides. This gives us $-3 \tan^2 x = -1$. Now, we have the term with $\tan^2 x$ almost isolated. The next step is to get rid of the coefficient, which is -3. To do this, we divide both sides of the equation by -3. So, dividing both sides by -3, we get $\tan^2 x = \frac{-1}{-3}$, which simplifies to $\tan^2 x = \frac{1}{3}$. Great! We've successfully isolated $\tan^2 x$. This sets us up perfectly for the next step, where we'll tackle the square and find $\tan x$.

Step 2: Take the Square Root

Now that we've isolated $\tan^2 x$, our next move is to get rid of that square. To do this, we'll take the square root of both sides of the equation. Remember, when you take the square root, you need to consider both the positive and negative roots. So, we have $\tan^2 x = \frac{1}{3}$. Taking the square root of both sides gives us $\sqrt{\tan^2 x} = \pm \sqrt{\frac{1}{3}}$. This simplifies to $\tan x = \pm \frac{1}{\sqrt{3}}$. It's super important to include both the positive and negative roots because the tangent function can be positive or negative depending on the quadrant. Now, let's rationalize the denominator. We have $\tan x = \pm \frac{1}{\sqrt{3}}$. To rationalize, we multiply the numerator and the denominator by $\sqrt{3}$, giving us $\tan x = \pm \frac{\sqrt{3}}{3}$. So, now we have two separate equations to solve: $\tan x = \frac{\sqrt{3}}{3}$ and $\tan x = -\frac{\sqrt{3}}{3}$. This sets us up perfectly for finding the reference angles and the solutions in the unit circle.

Step 3: Find the Reference Angles

Okay, so we've got $\tan x = \frac{\sqrt{3}}{3}$ and $\tan x = -\frac{\sqrt{3}}{3}$. Now, we need to find the reference angles for these equations. Reference angles are the acute angles formed by the terminal side of the angle and the x-axis. They help us find the solutions within the first rotation (0 to $2\pi$). For $\tan x = \frac{\sqrt{3}}{3}$, we need to think about which angle has a tangent of $\frac{\sqrt{3}}{3}$. If you recall your special triangles or the unit circle, you'll remember that $\tan \frac{\pi}{6} = \frac{\sqrt{3}}{3}$. So, the reference angle for this equation is $\frac{\pi}{6}$. Now, let's consider $\tan x = -\frac{\sqrt{3}}{3}$. The negative sign tells us that we're looking for angles where the tangent is negative. However, we still use the same reference angle, which is $\frac{\pi}{6}$, because we're just considering the magnitude of the tangent value for the reference angle. The negative sign will help us determine the quadrants where the solutions lie. So, in both cases, our reference angle is $\frac{\pi}{6}$. This is a crucial step because it simplifies finding the actual solutions in the different quadrants.

Step 4: Determine the Quadrants

Alright, we've nailed down the reference angle, which is $\frac{\pi}{6}$. Now, it's time to figure out which quadrants our solutions lie in. Remember, the tangent function is positive in the first and third quadrants and negative in the second and fourth quadrants. So, let's start with $\tan x = \frac{\sqrt{3}}{3}$. Since the tangent is positive, we're looking at the first and third quadrants. In the first quadrant, the angle is simply the reference angle, which is $\frac{\pi}{6}$. In the third quadrant, we find the angle by adding the reference angle to $\pi$. So, the angle in the third quadrant is $\pi + \frac{\pi}{6} = \frac{7\pi}{6}$. Now, let's tackle $\tan x = -\frac{\sqrt{3}}{3}$. Since the tangent is negative, we're looking at the second and fourth quadrants. In the second quadrant, we find the angle by subtracting the reference angle from $\pi$. So, the angle in the second quadrant is $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$. In the fourth quadrant, we find the angle by subtracting the reference angle from $2\pi$. So, the angle in the fourth quadrant is $2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$. So, we have four potential solutions: $\frac{\pi}{6}$, $\frac{7\pi}{6}$, $\frac{5\pi}{6}$, and $\frac{11\pi}{6}$. But remember, we need to express the general solutions.

Step 5: Express the General Solutions

Okay, guys, we're almost there! We've found the solutions within one rotation, but now we need to express the general solutions. This means we need to account for all possible solutions by adding integer multiples of the period of the tangent function. The period of the tangent function is $\pi$, which means the function repeats itself every $\pi$ radians. So, for $\tan x = \frac{\sqrt{3}}{3}$, we have the solutions $\frac{\pi}{6}$ and $\frac{7\pi}{6}$. To express the general solutions, we add $k\pi$ to each of these, where k is an integer. This gives us $\frac{\pi}{6} + k\pi$ and $\frac{7\pi}{6} + k\pi$. Notice that $\frac{7\pi}{6}$ is exactly $\pi$ more than $\frac{\pi}{6}$, so we can actually combine these into a single expression: $\frac{\pi}{6} + k\pi$. Now, let's consider $\tan x = -\frac{\sqrt{3}}{3}$. We have the solutions $\frac{5\pi}{6}$ and $\frac{11\pi}{6}$. Again, to express the general solutions, we add $k\pi$ to each of these, giving us $\frac{5\pi}{6} + k\pi$ and $\frac{11\pi}{6} + k\pi$. Similarly, $\frac{11\pi}{6}$ is exactly $\pi$ more than $\frac{5\pi}{6}$, so we can combine these into a single expression: $\frac{5\pi}{6} + k\pi$. So, our general solutions are $\frac{\pi}{6} + k\pi$ and $\frac{5\pi}{6} + k\pi$, where k is any integer. And that's it! We've found all the exact solutions for the given equation. High five!

Final Answer

Therefore, the exact solutions of the equation $-3 \tan^2 x + 1 = 0$ in radians are:

$\frac{\pi}{6} + k\pi$ and $\frac{5\pi}{6} + k\pi$, where k is an integer.

So, the correct answer is B. $\frac{\pi}{6}+k \pi$ and $\frac{5 \pi}{6}+k \pi$