Solving 2cos(3x) = 1: Find Solutions In [0, Π)

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Hey guys! Today, we're diving into a fun little trigonometric equation. Specifically, we're going to find all the solutions to the equation $2 \cos 3x = 1$ within the interval $[0, \pi)$. Ready? Let's jump right in!

Understanding the Problem

Before we start crunching numbers, let's make sure we understand what we're trying to do. We need to find all the values of x that make the equation $2 \cos 3x = 1$ true, but only the values that fall between 0 and $\pi$ (that's about 3.14159 radians, for those keeping score at home!).

Isolating the Cosine Function

The first thing we need to do is isolate the cosine function. That means getting $\cos 3x$ by itself on one side of the equation. To do this, we simply divide both sides of the equation by 2:

cos⁡3x=12\cos 3x = \frac{1}{2}

Okay, great! Now we have a simpler equation to work with. We know that the cosine of some angle is equal to $\frac1}{2}$. The big question is what angles have a cosine of $\frac{1{2}$?

Finding the General Solutions

Think back to your unit circle, or that handy cosine graph you have memorized (you do have it memorized, right?). We know that $\cos \theta = \frac{1}{2}$ when $\theta = \frac{\pi}{3}$ (that's 60 degrees) and $\theta = \frac{5\pi}{3}$ (that's 300 degrees). But hold on, there's more! Cosine is a periodic function, meaning it repeats itself every $2\pi$ radians. So, we can add any multiple of $2\pi$ to these angles and still get a cosine of $\frac{1}{2}$.

Therefore, the general solutions for $3x$ are:

3x=π3+2nπ3x = \frac{\pi}{3} + 2n\pi

and

3x=5π3+2nπ3x = \frac{5\pi}{3} + 2n\pi

where n is any integer (..., -2, -1, 0, 1, 2, ...).

Solving for x

Alright, we're almost there! Now we need to solve for x in each of these equations. To do that, we simply divide both sides of each equation by 3:

x=π9+2nπ3x = \frac{\pi}{9} + \frac{2n\pi}{3}

and

x=5π9+2nπ3x = \frac{5\pi}{9} + \frac{2n\pi}{3}

These are the general solutions for x. But remember, we're only interested in the solutions that fall within the interval $[0, \pi)$.

Finding Solutions in the Interval [0, π)

Now comes the fun part: plugging in different values of n to find the solutions that fit our interval.

Analyzing the First General Solution

Let's start with the first general solution:

x=π9+2nπ3x = \frac{\pi}{9} + \frac{2n\pi}{3}

  • n = 0: $x = \frac{\pi}{9} + \frac{2(0)\pi}{3} = \frac{\pi}{9}$. This is approximately 0.349 radians, which is definitely within our interval. So, $\frac{\pi}{9}$ is one of our solutions!
  • n = 1: $x = \frac{\pi}{9} + \frac{2(1)\pi}{3} = \frac{\pi}{9} + \frac{6\pi}{9} = \frac{7\pi}{9}$. This is approximately 2.443 radians, which is also within our interval. So, $\frac{7\pi}{9}$ is another solution!
  • n = 2: $x = \frac{\pi}{9} + \frac{2(2)\pi}{3} = \frac{\pi}{9} + \frac{12\pi}{9} = \frac{13\pi}{9}$. This is approximately 4.538 radians, which is outside our interval (since it's greater than $\pi$). So, we can stop increasing n for this solution, because all larger values of n will also be outside the interval.
  • n = -1: $x = \frac{\pi}{9} + \frac{2(-1)\pi}{3} = \frac{\pi}{9} - \frac{6\pi}{9} = -\frac{5\pi}{9}$. This is approximately -1.745 radians, so is outside of our interval and we need not look at values of n less than -1 either.

Analyzing the Second General Solution

Now let's look at the second general solution:

x=5π9+2nπ3x = \frac{5\pi}{9} + \frac{2n\pi}{3}

  • n = 0: $x = \frac{5\pi}{9} + \frac{2(0)\pi}{3} = \frac{5\pi}{9}$. This is approximately 1.745 radians, which is within our interval. So, $\frac{5\pi}{9}$ is a solution!
  • n = 1: $x = \frac{5\pi}{9} + \frac{2(1)\pi}{3} = \frac{5\pi}{9} + \frac{6\pi}{9} = \frac{11\pi}{9}$. This is approximately 3.840 radians, which is outside our interval (since it's greater than $\pi$). So, we can stop increasing n for this solution.
  • n = -1: $x = \frac{5\pi}{9} + \frac{2(-1)\pi}{3} = \frac{5\pi}{9} - \frac{6\pi}{9} = -\frac{\pi}{9}$. This is approximately -0.349 radians, so is outside of our interval and we need not look at values of n less than -1 either.

The Solutions

Alright, we've done it! We've found all the solutions to the equation $2 \cos 3x = 1$ in the interval $[0, \pi)$. They are:

  • x=π9x = \frac{\pi}{9}

  • x=5π9x = \frac{5\pi}{9}

  • x=7π9x = \frac{7\pi}{9}

So, those are our three solutions! You can always check your work by plugging these values back into the original equation to make sure they work.

Quick Recap

Here's a quick rundown of what we did:

  1. Isolate the cosine function: We rewrote the equation as $\cos 3x = \frac{1}{2}$.
  2. Find the general solutions: We found the general solutions for $3x$ using the unit circle and the periodicity of the cosine function.
  3. Solve for x: We divided by 3 to find the general solutions for x.
  4. Find solutions in the interval: We plugged in different values of n to find the solutions that fell within the interval $[0, \pi)$.

Hope this helps you master these types of problems! Keep practicing, and you'll be a trig whiz in no time!

Remember to always double-check your answers and make sure they make sense in the context of the problem.