Solving 2 Cos³x - 3 Cos²x + 1 = 0: A Trigonometric Guide
Hey guys! Today, we're diving deep into solving a trigonometric equation: 2 cos³x - 3 cos²x + 1 = 0. This might look intimidating at first, but don't worry, we'll break it down step-by-step. We’ll explore various techniques and strategies to tackle this problem, ensuring you understand not just the solution, but the process behind it. Whether you're a student prepping for an exam or just a math enthusiast, this guide is designed to help you master trigonometric equations. Let's jump right in and make some math magic happen!
Understanding the Equation
Before we start crunching numbers, let's understand the equation: 2 cos³x - 3 cos²x + 1 = 0. This is a cubic equation in terms of cos x. Recognizing this form is crucial because it allows us to use algebraic techniques we already know. Think of it like this: if we replace cos x with a simple variable, say 'y', the equation transforms into 2y³ - 3y² + 1 = 0. Suddenly, it looks much more familiar, right? This transformation is a powerful tool in solving trigonometric equations because it bridges the gap between trigonometry and algebra. We can apply methods like factoring or the rational root theorem to find solutions for 'y', and then convert back to cos x to solve for x. So, the first step in tackling such equations is often to make this mental or actual substitution to simplify the problem. This approach not only makes the equation less daunting but also opens the door to a variety of solution strategies that might not be immediately obvious when looking at the trigonometric form.
Initial Assessment
When you first encounter an equation like 2 cos³x - 3 cos²x + 1 = 0, it's crucial to take a moment for initial assessment. This involves recognizing the type of equation it is—in this case, a cubic equation in terms of cos x. Identifying this immediately helps you strategize your approach. For instance, knowing it's a cubic equation suggests that there might be up to three solutions, and techniques like factoring or using the rational root theorem could be applicable. Another key aspect of the initial assessment is looking for any immediate simplifications or substitutions that can be made. As we discussed earlier, replacing cos x with a single variable, like 'y', can drastically simplify the equation's appearance and make it more manageable. Furthermore, consider the range of possible values for cos x, which is always between -1 and 1. This constraint can be incredibly useful in narrowing down potential solutions later on. By carefully assessing these elements at the outset, you set the stage for a more efficient and accurate problem-solving process. This initial step prevents you from diving in blindly and helps you chart a clear path to the solution.
Solving the Cubic Equation
Now, let's get our hands dirty and solve the cubic equation. As we’ve discussed, the equation 2 cos³x - 3 cos²x + 1 = 0 can be viewed as a polynomial equation in cos x. To solve it, we can use techniques like factoring or the rational root theorem. Factoring, if possible, is often the quickest route. We look for factors of the constant term (1) that, when plugged into the equation, make it equal to zero. In this case, cos x = 1 is a clear solution because 2(1)³ - 3(1)² + 1 = 0. This tells us that (cos x - 1) is a factor of the cubic polynomial. Now, we can perform polynomial division or synthetic division to find the remaining quadratic factor. This process breaks down the cubic equation into simpler parts, making it much easier to handle. Once we have the factored form, we can set each factor equal to zero and solve for cos x. This typically involves solving a linear equation (from the (cos x - 1) factor) and a quadratic equation. Remember, the goal here is to find all possible values of cos x that satisfy the equation, and then use these values to find the corresponding values of x.
Factoring the Equation
Factoring the equation 2 cos³x - 3 cos²x + 1 = 0 is a crucial step in finding its solutions. As we identified, cos x = 1 is a root of the equation, meaning (cos x - 1) is a factor. To find the other factor, we can perform polynomial division. Divide 2 cos³x - 3 cos²x + 1 by (cos x - 1). This process will yield a quadratic expression. Alternatively, you can use synthetic division, which is a more streamlined method for dividing polynomials, especially when the divisor is a linear factor. Once you've performed the division, you'll have the equation in a factored form, something like (cos x - 1)(quadratic expression) = 0. The next step is to solve the quadratic expression. This might involve factoring it further, if possible, or using the quadratic formula. Factoring is often the more elegant approach if the quadratic expression can be easily factored. However, the quadratic formula is a reliable method that always works, regardless of whether the quadratic can be factored easily. By successfully factoring the original cubic equation, we transform it into simpler, more manageable equations that we can solve individually.
Applying the Rational Root Theorem
If factoring doesn't immediately jump out at you, the rational root theorem is your next best friend. This theorem is especially useful for polynomial equations with integer coefficients, like our 2 cos³x - 3 cos²x + 1 = 0. The rational root theorem helps us identify potential rational roots (roots that can be expressed as a fraction) of the polynomial. It states that any rational root of the polynomial must be of the form ±(factor of the constant term) / (factor of the leading coefficient). In our case, the constant term is 1 and the leading coefficient is 2. So, the possible rational roots are ±1 and ±1/2. Now, we can test these potential roots by plugging them into the equation to see if they satisfy it. This is a bit of trial and error, but it narrows down the possibilities significantly. Once you find a root, say cos x = r, you know that (cos x - r) is a factor of the polynomial, and you can proceed with polynomial division or synthetic division as we discussed earlier. The rational root theorem is a powerful tool for cracking cubic and higher-degree polynomial equations, giving you a systematic way to find at least one root, which then opens the door to further factorization and solving the equation.
Finding the Values of cos x
After factoring or using the rational root theorem, we arrive at the crucial step of finding the values of cos x. Let's say we've successfully factored our equation 2 cos³x - 3 cos²x + 1 = 0 into something like (cos x - 1)(2 cos²x - cos x - 1) = 0. Now, we set each factor equal to zero and solve for cos x. The first factor, cos x - 1 = 0, gives us cos x = 1. For the quadratic factor, 2 cos²x - cos x - 1 = 0, we can either factor it further or use the quadratic formula. If it factors, we'll get two more values for cos x. If not, the quadratic formula will provide us with the solutions. Remember, the quadratic formula is cos x = [-b ± √(b² - 4ac)] / (2a), where a, b, and c are the coefficients of the quadratic equation. Once we've applied the quadratic formula (if necessary), we'll have all the possible values of cos x that satisfy the equation. However, it's essential to remember that the range of cos x is between -1 and 1. So, any values outside this range are not valid solutions. This step is a critical checkpoint in the problem-solving process, ensuring that we only consider solutions that are mathematically feasible.
Solving for cos x = 1
Solving for cos x = 1 is a straightforward but essential step in our process. When we have cos x = 1, we need to find all the angles x where the cosine function equals 1. Recall the unit circle: cosine corresponds to the x-coordinate of a point on the circle. Cosine is 1 at 0 radians (or 0 degrees), and it repeats every 2π radians (or 360 degrees). Therefore, the general solution for cos x = 1 is x = 2πn, where n is any integer. This means x can be 0, 2π, -2π, 4π, -4π, and so on. These are all the angles where the cosine function hits its maximum value of 1. When solving trigonometric equations, it’s crucial to consider the periodic nature of trigonometric functions. This periodicity means that there are infinitely many solutions, but they follow a predictable pattern. So, when we state x = 2πn, we're capturing all these solutions in a concise and complete form. This understanding is fundamental not just for this specific equation but for tackling a wide range of trigonometric problems. Always keep the unit circle in mind, as it's your best friend for visualizing and solving trigonometric equations!
Solving the Quadratic Factor
Now, let's tackle the quadratic factor. Suppose after our initial factoring, we're left with a quadratic equation in the form of 2 cos²x - cos x - 1 = 0. We have a couple of options here: factoring or using the quadratic formula. Factoring is the quicker route if you can spot the factors easily. In this case, the quadratic expression can be factored into (2 cos x + 1)(cos x - 1) = 0. If factoring isn't immediately obvious, the quadratic formula is your reliable backup. Remember, the quadratic formula is cos x = [-b ± √(b² - 4ac)] / (2a). Applying it to our equation, where a = 2, b = -1, and c = -1, will give us the solutions for cos x. Whether you factor or use the quadratic formula, you'll end up with two possible values for cos x. For each value, you'll need to find the corresponding angles x that satisfy the equation. Keep in mind the range of cosine, which is between -1 and 1. If any calculated value falls outside this range, it’s not a valid solution. This step ensures that we only consider realistic values for cos x, making our final solutions accurate and meaningful.
Finding the Values of x
With the values of cos x in hand, the next step is to find the corresponding values of x. This is where our knowledge of inverse trigonometric functions and the unit circle truly shines. For each value of cos x, we'll use the inverse cosine function, denoted as arccos or cos⁻¹, to find the principal angle. The principal angle is the angle within the range of 0 to π (0 to 180 degrees) whose cosine is the value we have. However, remember that the cosine function is positive in both the first and fourth quadrants, and negative in the second and third quadrants. This means that for each value of cos x, there are generally two angles within the interval [0, 2π) that satisfy the equation. To find all solutions, we need to consider the periodic nature of the cosine function. The cosine function has a period of 2π, so we can add or subtract multiples of 2π to our initial solutions to find all possible values of x. This is often expressed as x = θ + 2πn, where θ is the principal angle and n is an integer. By carefully considering the quadrants and the periodicity of the cosine function, we can ensure that we find all the solutions for x that satisfy our original equation.
Using Inverse Cosine
To find the values of x, we employ the inverse cosine function, often written as arccos or cos⁻¹. This function essentially undoes the cosine function, giving us the angle whose cosine is a given value. For instance, if we found that cos x = 0.5, then x = arccos(0.5). Your calculator will typically give you the principal value, which lies in the range of 0 to π (0 to 180 degrees). However, this is just one solution. Remember the unit circle? Cosine is positive in both the first and fourth quadrants. So, if arccos(0.5) gives you an angle in the first quadrant, there's another angle in the fourth quadrant with the same cosine value. To find it, you can subtract the principal value from 2π (or 360 degrees). This gives you the other angle within the 0 to 2π range. Similarly, if cos x is negative, the principal value will be in the second quadrant, and you can find the corresponding angle in the third quadrant. This careful consideration of quadrants ensures we capture all possible solutions within one period of the cosine function. Using the inverse cosine function is a crucial step, but understanding its limitations and the broader context of the unit circle is what truly unlocks the full solution set for our trigonometric equation.
Considering All Solutions
When solving trigonometric equations, it's absolutely crucial to consider all possible solutions. This is because trigonometric functions are periodic, meaning they repeat their values at regular intervals. We've already talked about finding the principal values using the inverse cosine function and identifying solutions within the interval [0, 2π). But the cosine function repeats every 2π radians. This means that if x is a solution, then x + 2π, x - 2π, x + 4π, and so on, are also solutions. To express this mathematically, we add 2πn to our principal solutions, where n is any integer. This gives us a general solution that covers all possible values of x. For example, if one solution is x = π/3, the general solution would be x = π/3 + 2πn. Similarly, if another solution is x = 5π/3, the general solution would be x = 5π/3 + 2πn. It's this addition of 2πn that captures the infinite number of solutions that arise from the periodic nature of the cosine function. Failing to consider this periodicity will lead to incomplete solutions, and you'll miss out on many valid answers. So, always remember to add that 2πn to each solution to fully capture the solution set of your trigonometric equation.
Verifying the Solutions
Before we declare victory, let's verify our solutions. This is a critical step in any mathematical problem-solving process, but especially so with trigonometric equations, where it's easy to introduce extraneous solutions due to the nature of the functions and the algebraic manipulations involved. To verify, simply plug each solution you've found back into the original equation, 2 cos³x - 3 cos²x + 1 = 0. If the equation holds true for that value of x, then it's a valid solution. If not, it's an extraneous solution that we need to discard. This check ensures that our solutions not only satisfy the factored forms or intermediate equations we worked with but also the original equation we set out to solve. It's like a final quality control step, catching any errors that might have crept in along the way. Verifying solutions can seem tedious, especially when you have multiple solutions to check, but it's a non-negotiable part of the problem-solving process. It gives you confidence in your answers and guarantees that you're presenting the correct solutions. So, make it a habit to always verify your solutions – it's the mark of a meticulous and accurate mathematician!
Plugging Solutions Back In
The most straightforward way to verify our solutions is by plugging them back into the original equation: 2 cos³x - 3 cos²x + 1 = 0. This process is simple: for each value of x you've found, calculate cos x, then substitute it into the equation. Evaluate both sides of the equation. If they are equal, then x is indeed a solution. If they are not, then x is an extraneous solution and should be discarded. For example, if one of our solutions was x = 0, we would calculate cos(0) = 1 and substitute it into the equation: 2(1)³ - 3(1)² + 1 = 2 - 3 + 1 = 0. Since this is true, x = 0 is a valid solution. However, if plugging in another value, say x = π, resulted in a non-zero outcome, we would know that π is not a valid solution for this equation. This process of plugging solutions back in is a direct and reliable way to confirm their validity. It leaves no room for ambiguity and gives you a clear yes or no answer for each potential solution. So, embrace this step as your final seal of approval on your hard work!
Conclusion
Alright guys, we've made it to the end! Solving the trigonometric equation 2 cos³x - 3 cos²x + 1 = 0 might have seemed daunting at first, but we’ve broken it down into manageable steps. We started by understanding the equation and recognizing it as a cubic in terms of cos x. Then, we used factoring and the rational root theorem to find the values of cos x. From there, we employed the inverse cosine function and considered the periodic nature of cosine to find all possible values of x. Finally, we verified our solutions by plugging them back into the original equation. This step-by-step approach is key to tackling any complex trigonometric equation. Remember, practice makes perfect! The more you work through problems like this, the more comfortable you'll become with the techniques and strategies involved. So, keep practicing, keep exploring, and don't be afraid to tackle those challenging equations. You've got this!