Mastering Double-Angle Formulas: Solve `cos(2θ)+3cos(θ)=4`

Hey there, math adventurers! Ever stared at a trigonometric equation and felt like it was speaking a different language? You're definitely not alone. Today, we're diving deep into solving a super common type of trigonometric equation: cos(2θ) + 3cos(θ) = 4 within the friendly confines of the interval [0, 2π). This isn't just about finding an answer; it's about understanding the tools at our disposal, especially those powerful double-angle formulas. We'll break down every step, make it feel approachable, and show you exactly how to conquer these kinds of problems with confidence. So, grab your imaginary protractor and let's get mathematical!

Understanding the Challenge: cos(2θ) + 3cos(θ) = 4

Alright, guys, let's talk about the beast we're taming today: cos(2θ) + 3cos(θ) = 4. At first glance, this equation might look a bit intimidating because we've got two different angle arguments – 2θ and θ. This is the main hurdle we need to overcome. You see, when you're trying to solve a trigonometric equation, it's usually much, much easier if all your terms involve the same angle. Having cos(2θ) and cos(θ) in the same equation is like trying to have a conversation with someone who's speaking in two different languages at the same time – confusing, right? Our primary goal is to transform this equation so that every trigonometric term refers to just θ. And guess what? This is where our trusty double-angle identity for cosine comes into play. It's truly the key to unlocking this puzzle.

Now, there are actually three main forms for the cos(2θ) identity, and knowing which one to pick is a bit of an art, but mostly logic. Let's list 'em out:

1. cos(2θ) = cos²(θ) - sin²(θ)
2. cos(2θ) = 2cos²(θ) - 1
3. cos(2θ) = 1 - 2sin²(θ)

So, which one do we choose for cos(2θ) + 3cos(θ) = 4? Well, if you look closely at our original equation, you'll notice that the other trigonometric term is 3cos(θ). This gives us a massive hint! If we can replace cos(2θ) with an expression that only involves cos(θ), we'd be in business. The second identity, cos(2θ) = 2cos²(θ) - 1, is our golden ticket! It allows us to express cos(2θ) entirely in terms of cos(θ), eliminating sin(θ) completely and getting everything aligned to θ. If we chose the first identity, we'd still have sin²(θ), which would then need another identity (sin²(θ) = 1 - cos²(θ)) to convert it anyway. The third one would leave us with sin(θ) as the primary function, which isn't ideal when the other term is cos(θ). By picking 2cos²(θ) - 1, we're setting ourselves up for success, aiming to turn this complex trigonometric equation into a simpler, more familiar quadratic equation. Once it's in a quadratic form, we can use our standard algebraic techniques like factoring or the quadratic formula, and then simply substitute back to find our angles. It's a fantastic strategy that's super common in advanced trigonometry, so understanding this initial step is absolutely crucial for mastering these equations. Keep this thought process in mind, as selecting the right identity is often the most critical first move!

Step-by-Step Solution: Unpacking the Double-Angle Formula

Alright, team, let's get down to brass tacks and actually solve this thing! Our equation, as we know, is cos(2θ) + 3cos(θ) = 4. Based on our previous discussion, the first and most critical step is to substitute the chosen double-angle identity into the equation. We're replacing cos(2θ) with 2cos²(θ) - 1. So, our equation transforms beautifully into:

(2cos²(θ) - 1) + 3cos(θ) = 4

See how that immediately makes things look a lot more manageable? Now all our terms involve cos(θ). The next move is to rearrange this equation to make it look like a standard quadratic equation, which, if you recall, is usually in the form ax² + bx + c = 0. To do this, we need to gather all terms on one side and set the equation to zero. Let's do that:

2cos²(θ) + 3cos(θ) - 1 - 4 = 0 2cos²(θ) + 3cos(θ) - 5 = 0

Voila! We have a beautiful quadratic equation in terms of cos(θ). Now, to make it even more familiar, let's employ a common substitution technique. We can temporarily let x = cos(θ). This substitution isn't strictly necessary, but it often helps people visualize and solve the quadratic more easily, especially if quadratics are your jam. So, with x = cos(θ), our equation becomes:

2x² + 3x - 5 = 0

Now, this is a plain old quadratic equation, and we have a few ways to solve it: factoring, completing the square, or using the quadratic formula. For this one, factoring looks like a pretty sweet deal. We're looking for two numbers that multiply to (2 * -5) = -10 and add up to 3. Those numbers are 5 and -2! So, we can rewrite the middle term and factor by grouping:

2x² + 5x - 2x - 5 = 0 x(2x + 5) - 1(2x + 5) = 0 (2x + 5)(x - 1) = 0

This gives us two possible solutions for x:

1. 2x + 5 = 0 => 2x = -5 => x = -5/2
2. x - 1 = 0 => x = 1

Now, here's the super important part, guys: remember that x was just a placeholder for cos(θ). So, we substitute cos(θ) back in:

1. cos(θ) = -5/2
2. cos(θ) = 1

We need to immediately check the validity of these solutions. Think back to what you know about the cosine function. The range of cos(θ) is always between -1 and 1, inclusive. That means cos(θ) can never be less than -1 or greater than 1. Looking at our first solution, cos(θ) = -5/2, which is -2.5. This value is outside the valid range for cosine! This means cos(θ) = -5/2 yields no actual solutions for θ. We can simply discard this one. It's a common trap, so always remember to check the range! Our only viable solution is cos(θ) = 1. This whole process of converting, simplifying, solving the quadratic, and then checking validity is a fundamental skill in solving complex trigonometric equations. Master these steps, and you're golden!

Finding θ: Navigating the Unit Circle and Inverse Cosine

Okay, so we've successfully whittled down our initial beast of an equation to a much simpler task: solving cos(θ) = 1. This is where our knowledge of the unit circle and the inverse cosine function really shines. Remember, we're looking for solutions for θ specifically within the interval [0, 2π). This interval means we include 0 but exclude 2π (because 0 and 2π represent the same angular position on the unit circle, and usually, we only list one unique angle in the solution set). So, let's find the angles where the cosine value is exactly 1.

If you recall your unit circle, the x-coordinate of any point on the circle represents the cosine of the angle. We're looking for where this x-coordinate is 1. This only happens at one specific point on the unit circle within the [0, 2π) interval: when θ = 0 radians. At θ = 0, the point on the unit circle is (1, 0), and indeed, cos(0) = 1. If we were to go around the circle again, we'd reach 2π, where cos(2π) = 1 as well, but since our interval is [0, 2π), we don't include 2π itself. So, θ = 0 is our single, unique solution for this particular problem within the specified interval. It's often surprising to folks that a seemingly complex equation can yield such a simple, elegant answer!

Let's take a moment to really emphasize the importance of the interval [0, 2π). If the problem asked for all possible solutions, we would express our answer using the general solution form: θ = 0 + 2nπ, where n is any integer. This 2nπ part accounts for the periodic nature of the cosine function – it repeats every 2π radians. However, since we're restricted to [0, 2π), we're essentially looking for the first positive rotation (or starting at zero) where our condition is met. Always double-check the interval specified in your problem, as it dictates how many solutions you should find and how you express them. Sometimes you might find two solutions in [0, 2π) (e.g., if cos(θ) = 1/2, you'd have θ = π/3 and θ = 5π/3), but in this case, cos(θ) = 1 is quite specific, only hitting 1 at the very beginning of our journey around the circle. Understanding the relationship between the unit circle, the trigonometric functions, their inverse functions, and the concept of periodicity is what truly solidifies your ability to solve these types of equations. Don't just memorize; visualize it on the unit circle! It makes a world of difference.

What About Half-Angle Formulas?

Now, the initial prompt for this problem mentioned both double-angle and half-angle formulas. While we only needed a double-angle formula for cos(2θ) + 3cos(θ) = 4, it's super important to understand what half-angle formulas are and when they come in handy. They're another powerful tool in your trig toolkit, even if they didn't get called into action for this specific battle. Think of them as the inverse operation to double-angle formulas, allowing you to find the sine, cosine, or tangent of half an angle if you know the trigonometric values of the full angle.

So, what do these formulas look like? Here are the most common ones:

• Cosine Half-Angle Formula: cos(α/2) = ±√[(1 + cos(α))/2]
• Sine Half-Angle Formula: sin(α/2) = ±√[(1 - cos(α))/2]
• Tangent Half-Angle Formulas:
  • tan(α/2) = ±√[(1 - cos(α))/(1 + cos(α))]
  • tan(α/2) = sin(α) / (1 + cos(α))
  • tan(α/2) = (1 - cos(α)) / sin(α)

Notice those ± signs? They're crucial! The sign depends entirely on the quadrant in which α/2 lies. You have to figure that out before you slap on a plus or minus. This is a common point where students can get tripped up, so always be mindful of the quadrant of the half-angle. For example, if α/2 is in the first quadrant, all values are positive. If α/2 is in the second quadrant, sine is positive, but cosine and tangent are negative.

When would you actually use these bad boys? Half-angle formulas are incredibly useful when you need to find the exact trigonometric values for angles that aren't