Finding The Smallest Positive Root Of A Trigonometric Equation

October 10, 2025 by Scholario Team 63 views

Hey guys! Let's dive into a fun algebra problem today. We're tackling a trigonometric equation and trying to find its smallest positive root. Sounds exciting, right? So, grab your thinking caps, and let's get started!

Understanding the Problem

First, let's break down the equation we're dealing with: $2\sin(31°+x) \cdot \sin (59° – x) = \cos5\pi$ . Our main goal here is to find the value of x that satisfies this equation, specifically the smallest positive value. This involves a mix of trigonometric identities and a bit of algebraic manipulation. Don't worry, we'll take it step by step!

The key here is recognizing the trigonometric functions and their relationships. We have sine functions with angle additions and subtractions, and a cosine function on the other side. To solve this, we'll need to use some trigonometric identities to simplify the equation. Remember, trigonometry is all about relationships between angles and sides of triangles, so understanding these relationships is crucial. We need to simplify the left-hand side of the equation, which involves the product of two sine functions. A useful identity here is the product-to-sum identity. This identity allows us to express the product of sine functions as a sum or difference of cosine functions, making it easier to handle. So, before diving into calculations, let's recall the product-to-sum identities. These identities are essential tools in solving trigonometric equations, and mastering them will make problems like these much easier to tackle. Remember, practice makes perfect, so the more you work with these identities, the more comfortable you'll become using them.

Trigonometric Identities to the Rescue

To solve this, we'll use the product-to-sum identity. Specifically, we need the identity that deals with the product of two sines. Remember this handy formula:

$2 \\sin A \\sin B = \\cos(A - B) - \\cos(A + B)$

This identity is going to be our best friend in simplifying the left side of the equation. It allows us to transform a product of sines into a difference of cosines, which is often easier to work with. Think of it like turning a multiplication problem into a subtraction problem – sometimes, that's exactly what you need to crack the code! Understanding and memorizing these identities is super important for any trigonometry problem. It's like having a Swiss Army knife for math – you never know when it might come in handy. Let's see how we can apply this identity to our specific equation and make it more manageable.

Applying the Identity

Let's identify our A and B in the equation $2\sin(31°+x) \cdot \sin (59° – x) = \cos5\pi$ .

A = 31° + x
B = 59° – x

Now, let's plug these into our identity:

$2 \\sin(31°+x) \\sin(59° – x) = \\cos((31° + x) - (59° – x)) - \\cos((31° + x) + (59° – x))$

This is where the magic happens! By substituting our A and B values into the identity, we're starting to unravel the complexity of the equation. It's like peeling back the layers of an onion, each step bringing us closer to the core solution. Now, let's simplify this expression further and see what we get. Remember, the key is to take things one step at a time and not rush the process. Math is like a puzzle, and each small simplification is a piece falling into place.

Simplifying the Equation

Let's simplify the cosine arguments:

\cos((31° + x) - (59° – x)) = \cos(31° + x - 59° + x) = \cos(2x - 28°)

\cos((31° + x) + (59° – x)) = \cos(31° + x + 59° – x) = \cos(90°)

So our equation now looks like this:

\cos(2x - 28°) - \cos(90°) = \cos(5\pi)

Awesome! We've made some serious progress here. By simplifying the cosine arguments, we've reduced the complexity of the equation even further. Notice how the x terms canceled out in one of the cosine arguments, leaving us with just \cos(90°). This is a classic example of how trigonometric identities can help us simplify seemingly complicated expressions. Now, let's evaluate \cos(90°) and \cos(5\pi) and see what we get.

Evaluating Cosine Values

We know that \cos(90°) = 0. Also, \cos(5\pi) = \cos(\pi + 4\pi) = \cos(\pi) = -1. Remember, cosine is negative in the second and third quadrants, and ${5\pi}$ lands on the negative x-axis, just like ${\pi}$ . So, our equation becomes:

\cos(2x - 28°) - 0 = -1

\cos(2x - 28°) = -1

Fantastic! We've managed to simplify the equation to a point where it's much easier to solve. Knowing the values of cosine at key angles like 90° and multiples of π is crucial for these types of problems. It's like having your multiplication tables memorized – it makes the process so much smoother and faster. Now that we have a simple equation involving a single cosine function, we can move on to finding the values of x that satisfy it. Let's keep going!

Solving for x

Now, we need to find the angles for which the cosine is -1. We know that \cos(\pi) = -1, and cosine has a period of ${2\pi}$ , so the general solution is:

2x - 28° = (2n + 1)180°, where n is an integer.

This is a crucial step! We're using our knowledge of the cosine function's behavior to find all possible solutions. Remember that cosine is -1 at π, 3π, 5π, and so on, which can be represented by the general form (2n + 1)π. By setting our argument (2x - 28°) equal to this general form, we're capturing all the angles where cosine is -1. Now, let's solve for x and see what we get.

Isolating x

Let's solve for x:

2x = (2n + 1)180° + 28°

x = (2n + 1)90° + 14°

Great job! We've successfully isolated x and found a general solution that depends on the integer n. This means there are infinitely many solutions to our equation, but we're interested in finding the smallest positive one. By dividing both sides of the equation by 2, we've expressed x in terms of n. Now, to find the smallest positive solution, we need to figure out which value of n gives us the smallest positive x. Let's try plugging in some values for n and see what happens.

Finding the Smallest Positive Root

Now, we need to find the smallest positive value for x. Let's try different integer values for n:

For n = 0: x = (2(0) + 1)90° + 14° = 90° + 14° = 104°
For n = -1: x = (2(-1) + 1)90° + 14° = -90° + 14° = -76° (This is negative, so we don't want it.)

So, the smallest positive root is 104°.

Woohoo! We did it! By systematically plugging in values for n, we found the smallest positive solution for x. This is a great example of how understanding the general solution can help us pinpoint the specific solution we're looking for. Remember, the key is to think step by step and not be afraid to try different values. Now, let's summarize our findings and celebrate our success!

Conclusion

The smallest positive root of the equation $2\sin(31°+x) \cdot \sin (59° – x) = \cos5\pi$ is 104°.

Great job, guys! We tackled a challenging trigonometric equation and came out victorious. Remember, the key to solving these problems is to break them down into smaller steps, use trigonometric identities wisely, and think logically. Keep practicing, and you'll become a trigonometry master in no time!