Solving: Sum Of Roots For 2sin(2x) - 5sin(x) + 2 = 0
Hey guys! Let's dive into a fun math problem today where we'll find the sum of the roots for a trigonometric equation. This kind of problem might seem a bit daunting at first, but don't worry, we'll break it down step by step. Our main goal is to figure out the sum of the roots for the equation 2sin(2x) - 5sin(x) + 2 = 0 within the interval 0 ≤ x < 2π. So, grab your thinking caps, and let's get started!
Understanding the Problem
Before we jump into solving, let's make sure we understand what the problem is asking. We have a trigonometric equation that involves sine functions, and we need to find all the possible values of x that make this equation true within a specific range (0 to 2π). These values are the 'roots' of the equation. Once we find these roots, we simply add them up to get our final answer. This requires a solid grasp of trigonometric identities and algebraic manipulation. Let's break down the key components:
- Trigonometric Equation: The equation involves sine functions, specifically sin(2x) and sin(x). We'll need to use trigonometric identities to simplify and solve it.
- Roots: These are the solutions to the equation, the values of x that make the equation equal to zero.
- Interval 0 ≤ x < 2π: This tells us the range within which we need to find our solutions. It's one full revolution around the unit circle.
Keeping these points in mind, we can approach the problem strategically and methodically. Remember, math problems like these are like puzzles, and every piece of information helps us get closer to the solution.
Step 1: Simplify the Equation
Okay, so the first thing we need to do is simplify our equation. The presence of both sin(2x) and sin(x) makes it a bit tricky to handle directly. So, let’s use a trigonometric identity to rewrite sin(2x). Remember the double-angle identity? It states that sin(2x) = 2sin(x)cos(x). This identity is a game-changer because it allows us to express sin(2x) in terms of sin(x) and cos(x), which will help us in simplifying the overall equation.
Substituting this identity into our original equation, 2sin(2x) - 5sin(x) + 2 = 0, we get:
2(2sin(x)cos(x)) - 5sin(x) + 2 = 0
Which simplifies to:
4sin(x)cos(x) - 5sin(x) + 2 = 0
Now, our equation looks a bit more manageable. We have successfully replaced sin(2x) with an expression involving sin(x) and cos(x). However, we still have both sine and cosine terms, which can make solving the equation difficult. The next logical step is to try and express everything in terms of a single trigonometric function. This will allow us to potentially factor the equation or use other algebraic techniques to find the roots. So, let's move on to the next step where we'll aim to do just that.
Step 2: Express in Terms of sin(x)
Alright, let's take our simplified equation from the previous step: 4sin(x)cos(x) - 5sin(x) + 2 = 0. We need to get everything in terms of sin(x). To do this, we can use another fundamental trigonometric identity: the Pythagorean identity. This identity states that sin²(x) + cos²(x) = 1. We can rearrange this to express cos(x) in terms of sin(x).
From the Pythagorean identity, we get:
cos²(x) = 1 - sin²(x)
Taking the square root of both sides:
cos(x) = ±√(1 - sin²(x))
Now, this gives us two possible expressions for cos(x), a positive and a negative square root. We'll need to consider both possibilities when solving for x. For now, let's substitute this expression for cos(x) back into our equation:
4sin(x)(±√(1 - sin²(x))) - 5sin(x) + 2 = 0
This equation looks a bit messier now, with the square root. To make it easier to handle, let's introduce a substitution. Let's say y = sin(x). This will simplify the equation and allow us to work with it more easily.
Substituting y for sin(x), we get:
4y(±√(1 - y²)) - 5y + 2 = 0
Now, the equation looks a bit cleaner. We have a single variable, y, and we can proceed with solving for it. However, we still have that pesky square root to deal with. In the next step, we'll tackle this by isolating the square root term and squaring both sides of the equation. This will eliminate the square root but might introduce some extraneous solutions, which we'll need to check later.
Step 3: Solve for sin(x) using substitution
Okay, we've made some good progress. Our equation now looks like this: 4y(±√(1 - y²)) - 5y + 2 = 0, where y = sin(x). To get rid of the square root, we need to isolate the term containing it and then square both sides of the equation. Let's start by isolating the square root term:
4y(±√(1 - y²)) = 5y - 2
Now, we square both sides of the equation:
(4y(±√(1 - y²)))² = (5y - 2)²
This simplifies to:
16y²(1 - y²) = 25y² - 20y + 4
Expanding the left side, we get:
16y² - 16y⁴ = 25y² - 20y + 4
Now, let's rearrange the equation to get a polynomial equation in terms of y:
16y⁴ + 9y² - 20y + 4 = 0
This is a quartic equation (an equation with the highest power of the variable being 4). Solving quartic equations can be tricky, but in some cases, we can simplify them or find rational roots. Let’s rearrange our polynomial equation to set it to zero and look for possible factorization opportunities.
This quartic equation looks intimidating, but let's try to factor it. Sometimes, quartic equations can be factored into simpler quadratic equations. After some manipulation, we can rewrite the equation as:
(y - 2)(ay^3 + by^2 + cy + d) = 0
However, to simplify things further and inspired by the original equation, we will try a different approach. Going back to the quartic form:
16y⁴ + 9y² - 20y + 4 = 0
Let's try a substitution method to reduce the complexity. We know that y = sin(x), and we are expecting simple root values since trigonometric problems often have neat solutions. By trying rational root theorem or factoring techniques, we can observe that this quartic equation might be factored. Let's attempt factoring by grouping or synthetic division to see if we can find a rational root easily. Testing rational roots like y = 1/2, y = 2, etc., would be helpful. Specifically, y = 1/2 seems like a prospective root.
By attempting synthetic division or plugging y = 1/2 into the equation, we find that it does not neatly factor into an obvious simple form directly. This suggests we might need to reconsider our factoring approach or explore numerical methods if an analytical solution is elusive. However, there is a different approach we can undertake by going back a step before we squared the equation.
4y√(1 - y²) = 5y - 2
Instead of expanding to get a quartic equation, we can recognize this form and solve for y directly. This avoids complicating the problem into a high-degree polynomial which is often very difficult to factor. Let’s rearrange this and see if we can get something solvable.
To make the solution clearer, we can recognize that the original equation can lead to quadratic forms without needing to handle the quartic directly. We look for values that would make sense in the context of trigonometric functions where −1 ≤ y ≤ 1. Remember, if we backtrack to: 4y√(1 - y²) = 5y - 2, we could try squaring this more strategically, but this can still lead to complexity.
Instead, let's try going back even further and using our original transformed equation more deliberately:
4sin(x)cos(x) - 5sin(x) + 2 = 0
We'll explore different trigonometric methods to solve the problem efficiently in the next section. Let's see what clever tricks we can use!
Step 4: Solving the Simplified Trigonometric Equation
Okay, let's take a step back and look at our simplified equation again: 4sin(x)cos(x) - 5sin(x) + 2 = 0. We've tried a few algebraic manipulations, and while they gave us some insights, they also led to some complex equations. Sometimes, in math, it's good to try a different approach. In this case, let’s go back to basics and use trigonometric intuition to guide us.
We know that 2sin(x)cos(x) = sin(2x), so we can rewrite the first term as 2sin(2x). This doesn't seem to directly simplify the equation to a factorable form, but it does remind us of our original equation and might lead to a better substitution strategy.
Let’s consider again 4sin(x)cos(x) - 5sin(x) + 2 = 0, and rewrite it by using the double-angle formula for sine:
2(2sin(x)cos(x)) - 5sin(x) + 2 = 0
2sin(2x) - 5sin(x) + 2 = 0
At this point, we face a mix of sin(2x) and sin(x), which presents a hurdle. We need to relate sin(2x) back to sin(x) to solve effectively. Recalling the double-angle identity sin(2x) = 2sin(x)cos(x) is key.
Now, let's substitute sin(2x) back with 2sin(x)cos(x):
2(2sin(x)cos(x)) - 5sin(x) + 2 = 0
4sin(x)cos(x) - 5sin(x) + 2 = 0
This equation is tricky because we have both sine and cosine. To proceed, we can use the Pythagorean identity to express everything in terms of sine. Let's substitute cos(x) with ±√(1 - sin²(x)):
4sin(x)(±√(1 - sin²(x))) - 5sin(x) + 2 = 0
Now, substitute y = sin(x):
4y(±√(1 - y²)) - 5y + 2 = 0
We had this equation before! Let’s isolate the square root:
4y√(1 - y²) = 5y - 2
Square both sides:
16y²(1 - y²) = (5y - 2)²
16y² - 16y⁴ = 25y² - 20y + 4
Rearrange to a quartic equation:
16y⁴ + 9y² - 20y + 4 = 0
As we observed before, handling the quartic directly is challenging. Going back to our previous step before squaring might be beneficial. So, we will take the equation before squaring:
4y√(1 - y²) = 5y - 2
To avoid complexity, let’s try a different trigonometric substitution directly. Back to:
2sin(2x) - 5sin(x) + 2 = 0
We may try using trigonometric identities in a more targeted way. This time, consider the possibility that there are solutions where sin(x) equals 1/2 or 2. If sin(x) is 2, this is impossible since the range of sine is [-1, 1]. So, let's check for sin(x) = 1/2.
If sin(x) = 1/2, then x can be π/6 or 5π/6 in the interval [0, 2π). Now, we should verify if these are indeed the solutions by substituting them back into the original equation.
Let’s check x = π/6:
2sin(2(π/6)) - 5sin(π/6) + 2
2sin(π/3) - 5(1/2) + 2
2(√3/2) - 5/2 + 2
√3 - 5/2 + 2
√3 - 1/2 ≠ 0
So, π/6 is not a solution. Now, let's check x = 5π/6:
2sin(2(5π/6)) - 5sin(5π/6) + 2
2sin(5π/3) - 5(1/2) + 2
2(-√3/2) - 5/2 + 2
-√3 - 5/2 + 2
-√3 - 1/2 ≠ 0
Thus, 5π/6 is also not a direct solution. This indicates the solutions are not straightforward to guess, and we need a more systematic algebraic approach.
Step 5: Finding the Sum of the Roots
Given the challenges we've encountered, let's pivot and focus on finding the sum of the roots using Vieta's formulas, if applicable, or alternative trigonometric approaches. Backtracking to a more manageable form, we had:
2sin(2x) - 5sin(x) + 2 = 0
Let’s substitute sin(2x) = 2sin(x)cos(x) again:
4sin(x)cos(x) - 5sin(x) + 2 = 0
Now, express cos(x) in terms of sin(x) using cos(x) = ±√(1 - sin²(x)):
4sin(x)(±√(1 - sin²(x))) - 5sin(x) + 2 = 0
Let y = sin(x):
4y(±√(1 - y²)) - 5y + 2 = 0
Isolate the radical term:
4y√(1 - y²) = 5y - 2
Square both sides:
16y²(1 - y²) = (5y - 2)²
16y² - 16y⁴ = 25y² - 20y + 4
Rearrange to form a quartic equation:
16y⁴ + 9y² - 20y + 4 = 0
Finding the roots of this quartic equation is non-trivial and doesn't easily lend itself to Vieta's formulas directly because Vieta's formulas typically apply to polynomial equations in the variable x itself, not a function of x (like sin(x)).
However, we made an error in assuming a direct application of Vieta’s. To clarify the roots, we recognize that solving the quartic for y gives us values of sin(x), not x directly. Thus, summing the roots of the quartic doesn't give us the sum of x values that we need.
Considering trigonometric approach directly
Since directly solving the quartic and finding roots for y (which are sin(x) values) is complex, and Vieta's application isn't straightforward here, let's reconsider our original equation in its trigonometric form and look for possible algebraic tricks or insights.
2sin(2x) - 5sin(x) + 2 = 0
We realize that the equation mixes both sin(2x) and sin(x) terms. This is where we should apply trigonometric identities skillfully. We've tried the substitution approach leading to a quartic, and while logically sound, the algebraic complexity stalled progress. Let's backtrack to an even earlier form to reconsider our strategy. It’s important to step back and look at the forest for the trees.
Going back to the 4sin(x)cos(x) - 5sin(x) + 2 = 0 equation gives us an opportunity to attempt factoring or recognize patterns. However, a straightforward factorization doesn't appear obvious. Therefore, we will need to use the substitution wisely or look for a clever trick to relate x values that solve this equation.
Given this equation is complex to solve algebraically for all roots, recognizing the patterns to sum roots isn't straightforward either. We could potentially utilize numerical methods to approximate roots and sum them but without a clear analytical pathway, it poses a substantial challenge.
Let's Reconsider Substitution and Algebraic Manipulation
Given the difficulties in directly solving the quartic equation and the lack of an obvious trigonometric identity to simplify the sum of roots directly, let's revisit a critical step in our solution approach:
4sin(x)cos(x) - 5sin(x) + 2 = 0
We previously used the Pythagorean identity to express cos(x) in terms of sin(x). However, this led to complexities with the quartic equation. Instead of forcing everything into sine, let's make a crucial observation. If we let y = sin(x), the equation becomes:
4ycos(x) - 5y + 2 = 0
Isolating the cosine term gives us:
4ycos(x) = 5y - 2
If y ≠ 0, we can write:
cos(x) = (5y - 2) / (4y)
This is an interesting form, as it expresses cos(x) in terms of sin(x) directly. Since we also know that sin²(x) + cos²(x) = 1, we can substitute our expressions for sin(x) and cos(x) into this identity:
y² + ((5y - 2) / (4y))² = 1
Now, let's simplify:
y² + (25y² - 20y + 4) / (16y²) = 1
Multiply through by 16y² to clear the fraction (assuming y ≠ 0):
16y⁴ + 25y² - 20y + 4 = 16y²
Rearranging, we get our familiar quartic equation:
16y⁴ + 9y² - 20y + 4 = 0
As we've discussed, directly solving this quartic is challenging. The crux of this problem lies in a subtle approach rather than brute-force algebra.
Given the algebraic complexity and challenges encountered in directly solving the trigonometric equation or applying Vieta's formulas, it's indicative of a need for a more nuanced insight or a numerical method to approximate the sum of the roots. Analytical solution pathway complexity suggests there might be an error in expected outcomes without clear direct analytical steps. Given these challenges, a specific analytical answer isn't derivable within feasible computation steps from current manipulations due to complexity in deriving roots directly from quartic equations or simpler trigonometric relations. Therefore, we acknowledge the impasse in providing a precise sum of roots without further computational approximation or specific numerical techniques beyond direct application by algebraic transformations applied so far.
Conclusion
Solving trigonometric equations can be a fascinating journey, guys! In this case, we tackled a problem that seemed straightforward at first but quickly revealed its complexities. We learned about using trigonometric identities, algebraic substitutions, and the importance of strategic problem-solving. While we didn't arrive at a neat, easily calculable sum of the roots using elementary methods, we explored a range of techniques and gained a deeper understanding of the challenges involved in such problems. Remember, math isn't always about finding the perfect answer right away; it's about the process of exploration and learning that happens along the way. Keep practicing, keep exploring, and you'll become math whizzes in no time!