Mastering $\cos(x+\pi/6) = 1/2$: Solutions & Equivalent Equations

Hey there, math wizards and curious minds! Ever looked at a trigonometric equation and felt a tiny shiver of challenge? Well, you're in for a treat today because we're diving deep into a super common, yet incredibly important, type of problem: solving trigonometric equations, specifically one involving a cosine function with a phase shift. We're going to break down how to conquer $f(x)=\cos(x+\pi/6)=\frac{1}{2}$ on the interval $0 \leq x \leq 2\pi$, and then, we'll talk about how to spot other equations that might look different but actually share the exact same solutions. This isn't just about finding x; it's about understanding the heart of trigonometry and how these functions behave. We'll use a friendly, step-by-step approach, packed with tips and tricks to make sure you really grasp the concepts. By the end of this journey, you'll not only be a pro at solving this particular problem but also have a solid foundation for tackling more complex trigonometric challenges. So, grab your favorite beverage, settle in, and let's unlock the secrets of these fascinating mathematical expressions together! Our goal is to make this whole process feel natural and intuitive, demystifying any confusing bits and empowering you with the confidence to tackle any similar problem that comes your way. We'll explore the periodic nature of the cosine function, the impact of a phase shift, and the crucial role of the given interval in determining our final solutions. Understanding these elements is key to not just getting the right answer, but truly comprehending why it's the right answer. Ready to become a trigonometry superstar? Let's do this!

Understanding the Core Problem: $\cos(x+\pi/6) = 1/2$

Alright, guys, let's get down to business and really understand the main event: solving the trigonometric equation $f(x)=\cos(x+\pi/6)=\frac{1}{2}$ within the specified interval $0 \leq x \leq 2\pi$. This particular problem is a fantastic entry point into the world of advanced trigonometry because it combines a fundamental cosine equation with a crucial concept known as a phase shift. First off, let's break down the function itself: $f(x)=\cos(x+\pi/6)$. This isn't just a simple $\cos(x)$; the added $\pi/6$ inside the cosine argument shifts the entire graph horizontally. Knowing this is the first step to truly understanding the problem. The equation we're tasked with solving is when this function equals $\frac{1}{2}$. This means we're looking for all the x values between $0$ and $2\pi$ (inclusive) that make the cosine of $(x+\pi/6)$ equal to exactly one-half. To tackle this, a brilliant strategy is to use a substitution. Let's make things easier on ourselves by letting $\theta = x+\pi/6$. Now, our problem transforms into a more familiar one: $\cos(\theta) = \frac{1}{2}$. This is a classic equation that you've likely encountered before, and its solutions are fundamental to trigonometry. When does the cosine function equal $\frac{1}{2}$? Think about your unit circle or your special triangles! The primary angles where $\cos(\theta) = \frac{1}{2}$ are $\theta = \frac{\pi}{3}$ and $\theta = \frac{5\pi}{3}$. These are the values in the first and fourth quadrants, respectively, where the x-coordinate (which cosine represents) is positive one-half. However, remember that the cosine function is periodic, meaning it repeats its values every $2\pi$ radians. So, the general solutions for $\cos(\theta) = \frac{1}{2}$ are $\theta = \frac{\pi}{3} + 2n\pi$ and $\theta = \frac{5\pi}{3} + 2n\pi$, where n is any integer. This periodicity is super important because it means there could be multiple solutions within our given interval. Now that we have the general solutions for $\theta$, our next big step, and where many people sometimes get tripped up, is to relate these back to our original x variable and, crucially, to the specific interval $0 \leq x \leq 2\pi$. The interval for x will directly impact the valid range for our substituted \theta variable. Don't worry, we'll unpack that in the next section with even more clarity and practical examples. Just remember, the foundation of solving these equations lies in identifying those basic angles and understanding the periodic nature of trigonometric functions. You've got this!

Solving for X: Unpacking $x+\pi/6$

Okay, team, now that we've nailed down the values for $\theta$ where $\cos(\theta) = \frac{1}{2}$, it's time for the really important part: solving for our original variable, x, and making sure our solutions fit snugly within the specified interval $0 \leq x \leq 2\pi$. Remember, we made the substitution $\theta = x+\pi/6$. This means that for every $\theta$ solution we found, we need to set $x+\pi/6$ equal to it and then solve for x. But wait! Before we just plug and chug, we need to adjust our interval for \theta. If $0 \leq x \leq 2\pi$, then by adding $\pi/6$ to all parts of this inequality, we get the corresponding interval for $\theta$: $0 + \pi/6 \leq x+\pi/6 \leq 2\pi + \pi/6$. This simplifies to $\frac{\pi}{6} \leq \theta \leq \frac{13\pi}{6}$. This new interval for $\theta$ is absolutely critical because it tells us which of the general solutions for $\theta$ are actually relevant to our problem. Let's revisit our general solutions for $\theta$: $\theta = \frac{\pi}{3} + 2n\pi$ and $\theta = \frac{5\pi}{3} + 2n\pi$. Now, let's plug in different integer values for n and see which $\theta$ values fall within $\frac{\pi}{6} \leq \theta \leq \frac{13\pi}{6}$.

For the first set of solutions, $\theta = \frac{\pi}{3} + 2n\pi$:

• If $n=0$, then $\theta = \frac{\pi}{3}$. Is $\frac{\pi}{6} \leq \frac{\pi}{3} \leq \frac{13\pi}{6}$? Yes, because $\frac{\pi}{3}$ (which is $\frac{2\pi}{6}$) is comfortably between $\frac{\pi}{6}$ and $\frac{13\pi}{6}$.
• If $n=1$, then $\theta = \frac{\pi}{3} + 2\pi = \frac{\pi}{3} + \frac{6\pi}{3} = \frac{7\pi}{3}$. Is $\frac{\pi}{6} \leq \frac{7\pi}{3} \leq \frac{13\pi}{6}$? Let's convert $\frac{7\pi}{3}$ to sixths: $\frac{14\pi}{6}$. This is outside our interval of $\frac{13\pi}{6}$. So, $n=1$ is too large. If $n=-1$, then $\theta = \frac{\pi}{3} - 2\pi = -\frac{5\pi}{3}$, which is too small.

For the second set of solutions, $\theta = \frac{5\pi}{3} + 2n\pi$:

• If $n=0$, then $\theta = \frac{5\pi}{3}$. Is $\frac{\pi}{6} \leq \frac{5\pi}{3} \leq \frac{13\pi}{6}$? Yes, because $\frac{5\pi}{3}$ (which is $\frac{10\pi}{6}$) is between $\frac{\pi}{6}$ and $\frac{13\pi}{6}$.
• If $n=1$, then $\theta = \frac{5\pi}{3} + 2\pi = \frac{5\pi}{3} + \frac{6\pi}{3} = \frac{11\pi}{3}$. This is $\frac{22\pi}{6}$, which is also outside our interval. If $n=-1$, then $\theta = \frac{5\pi}{3} - 2\pi = -\frac{\pi}{3}$, which is too small.

So, the valid values for $\theta$ in our adjusted interval are just $\frac{\pi}{3}$ and $\frac{5\pi}{3}$. Awesome! Now, we can finally solve for x using these values:

1. For $\theta = \frac{\pi}{3}$: $x+\frac{\pi}{6} = \frac{\pi}{3}$. Subtract $\frac{\pi}{6}$ from both sides: $x = \frac{\pi}{3} - \frac{\pi}{6} = \frac{2\pi}{6} - \frac{\pi}{6} = \frac{\pi}{6}$.
2. For $\theta = \frac{5\pi}{3}$: $x+\frac{\pi}{6} = \frac{5\pi}{3}$. Subtract $\frac{\pi}{6}$ from both sides: $x = \frac{5\pi}{3} - \frac{\pi}{6} = \frac{10\pi}{6} - \frac{\pi}{6} = \frac{9\pi}{6} = \frac{3\pi}{2}$.

Let's double-check these solutions with our original interval for x, which was $0 \leq x \leq 2\pi$. Both $x=\frac{\pi}{6}$ and $x=\frac{3\pi}{2}$ fall perfectly within this range. So, there you have it! The solutions to $f(x)=\cos(x+\pi/6)=\frac{1}{2}$ on the interval $0 \leq x \leq 2\pi$ are $x=\frac{\pi}{6}$ and $x=\frac{3\pi}{2}$. Phew, that was a journey, but we made it! Understanding how to manipulate the interval for the substituted variable is arguably the most crucial part of solving these types of problems effectively. Keep up the great work!

Identifying Equivalent Equations: What Else Works?

Now, for the really intriguing part of our journey: identifying equivalent equations. The original problem asks us to find which other equations would yield the exact same solutions as $f(x)=\cos(x+\pi/6)=\frac{1}{2}$ on the interval $0 \leq x \leq 2\pi$. This means we're looking for different mathematical expressions that, when solved, give us precisely $x=\frac{\pi}{6}$ and $x=\frac{3\pi}{2}$, and no other solutions within that interval. This concept is fundamental to understanding algebraic and trigonometric manipulation, showing that there isn't always just one