Solve Cos(4θ) = 0.0596: Smallest 2 Solutions
Hey guys! Let's dive into a fun math problem today. We're going to find the smallest two solutions for the equation cos(4θ) = 0.0596 within the interval [0, 2π). This might sound a bit intimidating at first, but trust me, we'll break it down step by step and make it super clear. So, grab your thinking caps, and let's get started!
Understanding the Problem
Before we jump into the calculations, let's make sure we really understand what the question is asking. We're dealing with a trigonometric equation, specifically one involving the cosine function. Remember, the cosine function gives us the x-coordinate of a point on the unit circle. When we see cos(4θ), it means we're looking at the cosine of four times the angle θ. The value 0.0596 is a decimal, which means we are seeking angles whose cosine is very close to zero, thus, close to the y-axis on the unit circle. The interval [0, 2π) tells us that we're only interested in solutions that fall within one full revolution around the unit circle, starting from 0 radians and going up to (but not including) 2π radians. So, we need to find two angles, θ, that satisfy the equation and fall within this range. Make sense? Great! Now, let's get our hands dirty with the math.
Step-by-Step Solution
1. Finding the Principal Value
The first thing we need to do is find the principal value of 4θ. This is the angle whose cosine is 0.0596. We can use the inverse cosine function (also written as arccos or cos⁻¹) to find this value. So, we have:
4θ = arccos(0.0596)
Using a calculator (make sure it's in radian mode!), we find that:
4θ ≈ 1.5105 radians
This is our first solution for 4θ. But remember, the cosine function is positive in both the first and fourth quadrants. This means there's another angle in the interval [0, 2π) that also has a cosine of 0.0596. To find this angle, we can use the property that cos(x) = cos(2π - x). So, our second solution for 4θ is:
4θ ≈ 2π - 1.5105 ≈ 4.7727 radians
2. Finding All Solutions for 4θ
Now, we need to consider that the cosine function is periodic, meaning it repeats its values every 2π radians. This means that if we add multiples of 2π to our solutions for 4θ, we'll still get the same cosine value. So, the general solutions for 4θ are:
4θ ≈ 1.5105 + 2πk
4θ ≈ 4.7727 + 2πk
where k is any integer (..., -2, -1, 0, 1, 2, ...).
3. Solving for θ
Okay, we're getting closer! Now we need to actually find the values of θ, not just 4θ. To do this, we simply divide both sides of our general solutions by 4:
θ ≈ (1.5105 + 2πk) / 4
θ ≈ (4.7727 + 2πk) / 4
4. Finding Solutions in the Interval [0, 2π)
Now comes the crucial part: we need to find the solutions that fall within our interval [0, 2π). We'll do this by plugging in different integer values for k and seeing which solutions we get. Let's start with k = 0:
θ ≈ (1.5105 + 2π(0)) / 4 ≈ 0.3776 radians
θ ≈ (4.7727 + 2π(0)) / 4 ≈ 1.1932 radians
These are our first two solutions! They both fall within the interval [0, 2π). Now let's try k = 1:
θ ≈ (1.5105 + 2π(1)) / 4 ≈ 1.9475 radians
θ ≈ (4.7727 + 2π(1)) / 4 ≈ 2.7631 radians
These are also valid solutions within our interval. Let's try k = 2:
θ ≈ (1.5105 + 2π(2)) / 4 ≈ 3.5174 radians
θ ≈ (4.7727 + 2π(2)) / 4 ≈ 4.3330 radians
These are still within our interval. How about k = 3?
θ ≈ (1.5105 + 2π(3)) / 4 ≈ 5.0873 radians
θ ≈ (4.7727 + 2π(3)) / 4 ≈ 5.9029 radians
Still good! But what happens if we try k = 4?
θ ≈ (1.5105 + 2π(4)) / 4 ≈ 6.6572 radians
This is greater than 2π (which is approximately 6.2832), so it's outside our interval. We don't need to check any larger values of k. We also don't need to check negative values of k because those will give us negative solutions, which are also outside our interval. So, we have a total of eight solutions within the interval [0, 2π). But the question only asks for the smallest two! Looking at our list, the smallest two solutions are:
θ ≈ 0.3776 radians and θ ≈ 1.1932 radians
5. The Final Answer
Therefore, the smallest two solutions of cos(4θ) = 0.0596 on [0, 2π) are approximately 0.3776 and 1.1932. We can write this as:
0. 3776, 1.1932
Why This Matters: The Significance of Trigonometric Equations
Okay, so we've solved this equation, but you might be wondering,
