Solving 6sin²(w) - 7cos(w) - 8 = 0 Find All Solutions
Hey guys! Let's dive into solving this trigonometric equation together. We've got:
6sin²(w) - 7cos(w) - 8 = 0
for all solutions where 0 ≤ w < 2π. Trigonometric equations can seem daunting at first, but with a few key strategies, we can break them down and find the solutions. This article will guide you through the process step by step, ensuring you understand not only how to solve this equation but also why each step is necessary. We’ll explore the fundamental trigonometric identities, algebraic manipulations, and the unit circle to find all possible values of w within the given interval. So, grab your calculators, and let's get started!
1. Leveraging the Pythagorean Identity
The cornerstone of our approach lies in the Pythagorean identity: sin²(w) + cos²(w) = 1. This identity is crucial because it allows us to express sin²(w) in terms of cos²(w), or vice versa. In our equation, we have both sin²(w) and cos(w) terms. To simplify things, we'll replace sin²(w) with its equivalent expression derived from the Pythagorean identity. This substitution will transform our equation into a quadratic equation in terms of cos(w), which we can then solve using standard algebraic techniques. This is a common strategy when dealing with trigonometric equations involving both sine and cosine functions – aiming for a single trigonometric function makes the equation much easier to handle.
So, from the Pythagorean identity, we can write sin²(w) = 1 - cos²(w). Substituting this into our original equation gives us:
6(1 - cos²(w)) - 7cos(w) - 8 = 0
This substitution is a critical first step. By expressing the equation in terms of a single trigonometric function, we pave the way for algebraic manipulation and simplification. Without this step, we'd be stuck with an equation that's difficult to solve directly. The Pythagorean identity is a powerful tool in our trigonometric arsenal, and this is a perfect example of how it can be applied to solve complex equations. Remember, the key is to look for opportunities to use trigonometric identities to simplify the equation and make it more manageable. This foundational step will lead us closer to finding the solutions for w.
2. Transforming into a Quadratic Equation
Now that we've substituted using the Pythagorean identity, let's simplify the equation further. Expanding the terms, we get:
6 - 6cos²(w) - 7cos(w) - 8 = 0
Combining the constants, we have:
-6cos²(w) - 7cos(w) - 2 = 0
To make it look more familiar, we can multiply the entire equation by -1:
6cos²(w) + 7cos(w) + 2 = 0
Notice what we've got here, guys? This is a quadratic equation in terms of cos(w)! Think of cos(w) as our variable, say 'x'. The equation then looks like 6x² + 7x + 2 = 0. This transformation is a game-changer because we can now use our knowledge of quadratic equations to solve for cos(w). Quadratic equations are a fundamental part of algebra, and there are several methods we can use to find their roots, such as factoring, using the quadratic formula, or completing the square. In this case, factoring might be the most straightforward approach, but the quadratic formula will always work if factoring proves difficult. The ability to recognize and transform trigonometric equations into quadratic forms is a valuable skill in trigonometry, allowing us to apply familiar algebraic techniques to solve seemingly complex problems. This step is a testament to the interconnectedness of different branches of mathematics, where concepts from one area can be used to solve problems in another.
3. Solving the Quadratic Equation
Alright, let's tackle this quadratic equation: 6cos²(w) + 7cos(w) + 2 = 0. We're going to factor it. We need two numbers that multiply to (6 * 2 = 12) and add up to 7. Those numbers are 3 and 4. So, we can rewrite the middle term:
6cos²(w) + 3cos(w) + 4cos(w) + 2 = 0
Now, let's factor by grouping:
3cos(w)(2cos(w) + 1) + 2(2cos(w) + 1) = 0
Factor out the common term (2cos(w) + 1):
(3cos(w) + 2)(2cos(w) + 1) = 0
Now, we set each factor equal to zero:
3cos(w) + 2 = 0 or 2cos(w) + 1 = 0
Solving for cos(w) in each case:
cos(w) = -2/3 or cos(w) = -1/2
Great! We've found the values of cos(w) that satisfy our quadratic equation. But remember, we're not done yet. We need to find the actual values of w that correspond to these cosine values. This is where our understanding of the unit circle and inverse trigonometric functions comes into play. Each value of cos(w) corresponds to one or more angles within the interval 0 ≤ w < 2π. To find these angles, we'll use the inverse cosine function (arccos) and consider the quadrants where cosine is negative. This step is crucial because it connects the algebraic solution of the quadratic equation back to the trigonometric context of the original problem. We're essentially translating the values of cos(w) into angles, which are the solutions we're ultimately seeking.
4. Finding Solutions for w
Okay, we've got cos(w) = -2/3 and cos(w) = -1/2. Now, let's find the values of w within the interval 0 ≤ w < 2π.
Case 1: cos(w) = -2/3
Since cosine is negative, w lies in the second and third quadrants. We'll use the inverse cosine function to find the reference angle:
w_ref = arccos(2/3) ≈ 0.841 radians
In the second quadrant, w = π - w_ref ≈ π - 0.841 ≈ 2.301 radians
In the third quadrant, w = π + w_ref ≈ π + 0.841 ≈ 3.983 radians
Case 2: cos(w) = -1/2
We know that cos(π/3) = 1/2. Since cosine is negative, w lies in the second and third quadrants.
In the second quadrant, w = π - π/3 = 2π/3
In the third quadrant, w = π + π/3 = 4π/3
So, we have four potential solutions for w: approximately 2.301 radians, 3.983 radians, 2π/3, and 4π/3. It's essential to verify these solutions by plugging them back into the original equation to ensure they are valid. This step helps catch any extraneous solutions that might have arisen during the algebraic manipulations. Finding the solutions for w involves a combination of using inverse trigonometric functions, understanding the unit circle, and considering the quadrants where the trigonometric function has the correct sign. This process highlights the geometric interpretation of trigonometric functions and their relationship to angles.
5. Verifying the Solutions
It's super important to verify our solutions to make sure they actually work in the original equation and that we haven't made any mistakes along the way. This step is crucial because sometimes, when solving equations, we might end up with solutions that don't satisfy the original equation. These are called extraneous solutions, and they can arise from various algebraic manipulations, such as squaring both sides of an equation or, in this case, using trigonometric identities. To verify, we'll substitute each of our potential solutions back into the original equation: 6sin²(w) - 7cos(w) - 8 = 0. If the equation holds true for a particular value of w, then that value is a valid solution. If the equation does not hold true, then we discard that value as an extraneous solution.
Let's check our solutions:
- w ≈ 2.301 radians
- w ≈ 3.983 radians
- w = 2π/3
- w = 4π/3
After plugging these values back into the original equation, we can confirm that all four solutions are valid.
6. The Complete Solution Set
Awesome! After all that work, we've found our solutions and verified them. So, the solutions to the equation 6sin²(w) - 7cos(w) - 8 = 0 for 0 ≤ w < 2π are:
w ≈ 2.301 radians w ≈ 3.983 radians w = 2π/3 w = 4π/3
We can express these solutions as a set:
{2.301, 3.983, 2π/3, 4π/3}
And there you have it! We've successfully navigated through this trigonometric equation, using the Pythagorean identity, solving a quadratic equation, and understanding the unit circle. Remember, the key to solving trigonometric equations is to break them down into manageable parts, use trigonometric identities to simplify, and always verify your solutions. This comprehensive approach ensures that we not only find the solutions but also understand the underlying principles and concepts. Solving trigonometric equations is a valuable skill in mathematics, with applications in various fields such as physics, engineering, and computer science. So, keep practicing, and you'll become a pro at tackling these types of problems! Well done, guys!
