Solving Tan(x) = √3/3 A Step By Step Radians Guide

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Hey guys! Let's dive into solving a trigonometric equation today. We're going to figure out which radian value satisfies the equation $\tan x = \frac{\sqrt{3}}{3}$. This is a classic problem in trigonometry, and understanding how to solve it is super important for any math enthusiast. We'll break it down step by step, so don't worry if it seems a bit tricky at first.

Understanding the Tangent Function

First, let’s get a handle on what the tangent function actually represents. In a right-angled triangle, the tangent of an angle is the ratio of the length of the opposite side to the length of the adjacent side. Mathematically, we express it as:

tan⁡(θ)=OppositeAdjacent\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}

But there’s more to it than just triangles! Think about the unit circle, which is a circle with a radius of 1 centered at the origin in a coordinate plane. If you draw a line from the origin to a point on the circle, the angle this line makes with the positive x-axis is our angle θ\theta. The coordinates of that point on the unit circle are (cos⁡(θ),sin⁡(θ))(\cos(\theta), \sin(\theta)). Now, the tangent function in terms of the unit circle is:

tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}

This is crucial because it connects the tangent function to both sine and cosine, which helps us visualize and solve trigonometric equations. When we're looking for a solution to $ an x = \frac{\sqrt{3}}{3}$, we’re essentially searching for angles xx where the ratio of the sine to the cosine equals $\frac{\sqrt{3}}{3}$.

Visualizing Tangent

Imagine the unit circle. The tangent function repeats its values every π\pi radians because after half a rotation around the circle, the ratios of sine and cosine start repeating (with the same sign). This periodicity is a key characteristic of the tangent function. Think about how the tangent changes as you move around the circle. In the first quadrant (0 to $\frac{\pi}{2}$), both sine and cosine are positive, so tangent is also positive. As you move into the second quadrant ($\frac{\pi}{2}$ to $\pi$), sine is positive but cosine is negative, making tangent negative. This pattern continues, alternating between positive and negative values in different quadrants.

Understanding this behavior helps us narrow down the possible solutions. We know that $\frac{\sqrt{3}}{3}$ is a positive value, so we should be looking for angles where both sine and cosine have the same sign. This occurs in the first and third quadrants. By visualizing the unit circle and the tangent function’s behavior, we gain a much stronger intuition for solving trigonometric equations.

Identifying the Reference Angle

So, we’re trying to solve $\tan x = \frac{\sqrt{3}}{3}$. The first thing we need to do is find the reference angle. What’s a reference angle, you ask? Well, it’s the acute angle (an angle less than $\frac{\pi}{2}$ or 90 degrees) formed by the terminal side of the angle and the x-axis. Think of it as the angle within the first quadrant that helps us find our solution.

Now, we need to recall our special trigonometric values. You know, those common angles like 0, $\frac{\pi}{6}$, $\frac{\pi}{4}$, $\frac{\pi}{3}$, and $\frac{\pi}{2}$ (or 0°, 30°, 45°, 60°, and 90°) and their sine, cosine, and tangent values. These are super useful to memorize!

We know that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$. We’re looking for an angle where this ratio equals $\frac{\sqrt{3}}{3}$. Let’s look at some common values:

  • For $\theta = \frac{\pi}{6}$ (30°), $\sin(\frac{\pi}{6}) = \frac{1}{2}$ and $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$. So,

    tan⁡(π6)=1232=13=33\tan(\frac{\pi}{6}) = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}

    Bingo! That’s exactly what we’re looking for.

  • For $\theta = \frac{\pi}{3}$ (60°), $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ and $\cos(\frac{\pi}{3}) = \frac{1}{2}$. Thus,

    tan⁡(π3)=3212=3\tan(\frac{\pi}{3}) = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}

    Not quite!

So, the reference angle that satisfies our equation is $\frac{\pi}{6}$. But hold on, we're not done yet! We need to consider all possible solutions within the given range or general solutions if no range is specified.

The Importance of Special Trigonometric Values

Knowing these special trigonometric values off the top of your head makes solving these kinds of problems much faster. They act like building blocks for more complex trigonometry. Think of it like knowing your times tables in basic arithmetic. The more you practice and memorize these values, the easier trigonometric equations become!

Finding Solutions in All Quadrants

Okay, so we've found that the reference angle for $\tan x = \frac{\sqrt{3}}{3}$ is $\frac{\pi}{6}$. Awesome! But remember, the tangent function is positive in more than one quadrant. Tangent is positive in both the first and third quadrants. This is because tangent is the ratio of sine to cosine, and in the first quadrant, both sine and cosine are positive, while in the third quadrant, both are negative (a negative divided by a negative is a positive!).

So, we need to find the angles in both the first and third quadrants that have a reference angle of $\frac{\pi}{6}$.

  • First Quadrant: This one is straightforward. The angle is simply the reference angle itself, which is $\frac{\pi}{6}$.

  • Third Quadrant: To find the angle in the third quadrant, we add $\pi$ to our reference angle. This is because the third quadrant starts at $\pi$ radians, and we need to go an additional $\frac{\pi}{6}$ radians into that quadrant.

    x=π+π6=6π6+π6=7π6x = \pi + \frac{\pi}{6} = \frac{6\pi}{6} + \frac{\pi}{6} = \frac{7\pi}{6}

So, we have two solutions so far: $\frac{\pi}{6}$ and $\frac{7\pi}{6}$. These are the solutions within one full rotation around the unit circle (0 to 2$\pi$). If we weren't given a specific range, we'd need to consider all possible solutions, which means adding integer multiples of $\pi$ to these solutions, because the tangent function has a period of $\pi$. But for this problem, let's stick with the solutions within one rotation.

General Solutions

If the question asked for general solutions, we would express the solutions as:

x=π6+nπx = \frac{\pi}{6} + n\pi

where n is an integer. This accounts for all angles that have the same tangent value, as the tangent function repeats every $\pi$ radians.

The Correct Solution

Alright, we’ve done the hard work! We identified the reference angle, considered the quadrants where tangent is positive, and found two possible solutions: $\frac{\pi}{6}$ and $\frac{7\pi}{6}$. Now, let's take a look at the options given in the question.

The question asks for a solution, not all solutions. Among the options, we have:

  • π6\frac{\pi}{6}

  • π3\frac{\pi}{3}

We already know that $\frac{\pi}{6}$ is one of the solutions we found. Let's quickly check $\frac{\pi}{3}$ to be absolutely sure. We calculated earlier that $\tan(\frac{\pi}{3}) = \sqrt{3}$, which is not equal to $\frac{\sqrt{3}}{3}$. So, $\frac{\pi}{3}$ is not a solution.

Therefore, the correct solution is $\frac{\pi}{6}$.

Why Checking Your Answers Matters

It's always a good idea to double-check your work, especially in trigonometry. Plugging your solutions back into the original equation ensures that you haven't made any mistakes along the way. This is especially important when dealing with trigonometric equations because there can be multiple solutions, and it's easy to miss one or make a sign error.

Final Answer

So, there you have it! The solution to the equation $ an x = \frac{\sqrt{3}}{3}$ in radians is $\frac{\pi}{6}$. We've walked through the process step by step, from understanding the tangent function to finding the reference angle and considering all possible quadrants. I hope this explanation helps you tackle similar problems with confidence!

Answer: $\frac{\pi}{6}$