Solving Trigonometric Equations: A Step-by-Step Guide

Hey guys! Let's dive into solving the trigonometric equation $\sin^2(w) = -3\cos(w)$ for all solutions where $0 \leq w < 2\pi$. This might seem a little tricky at first, but trust me, we'll break it down into manageable steps. The goal here is to find all the values of 'w' (in radians) that make this equation true, and we need to give our answers accurate to two decimal places. Sounds good? Let's get started!

Step 1: Rewrite the Equation Using a Fundamental Identity

Our first move is to simplify the equation. Since we're dealing with $\sin^2(w)$ and $\cos(w)$, the Pythagorean identity is our best friend here. Remember that $\sin^2(w) + \cos^2(w) = 1$? We can rearrange this to express $\sin^2(w)$ in terms of $\cos^2(w)$: $\sin^2(w) = 1 - \cos^2(w)$. Now, substitute this into our original equation:

$1 - \cos^2(w) = -3\cos(w)$

This is great! We've got an equation with only $\cos(w)$ terms, which is a huge step forward. But it's still a bit messy. Let's rearrange it to look like a standard quadratic equation. Moving everything to one side, we get:

$\cos^2(w) - 3\cos(w) - 1 = 0$

See? Now it's starting to look familiar. This is a quadratic equation in terms of $\cos(w)$.

Why this step is crucial

This step is super important because it transforms our trigonometric equation into something we know how to solve. By using the Pythagorean identity, we've changed the equation to involve only one trigonometric function ($\cos(w)$). This allows us to treat the equation like a standard quadratic, making it much easier to solve. Without this transformation, we'd be stuck trying to solve an equation with two different trigonometric functions, which is far more complex. This is the key to unlocking the solution, so make sure you understand how to do it and why it works!

Step 2: Solve the Quadratic Equation

Alright, we've got a quadratic equation: $\cos^2(w) + 3\cos(w) - 1 = 0$. Now, let's solve for $\cos(w)$. Since this quadratic equation doesn't factor nicely, we'll use the quadratic formula. Remember the quadratic formula? For an equation in the form of $ax^2 + bx + c = 0$, the solutions are given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our case, we have $a = 1$, $b = 3$, and $c = -1$. Let's plug these values into the quadratic formula:

$\cos(w) = \frac{-3 \pm \sqrt{3^2 - 4(1)(-1)}}{2(1)}$

$\cos(w) = \frac{-3 \pm \sqrt{9 + 4}}{2}$

$\cos(w) = \frac{-3 \pm \sqrt{13}}{2}$

So, we have two possible values for $\cos(w)$:

$\cos(w) = \frac{-3 + \sqrt{13}}{2}$ and $\cos(w) = \frac{-3 - \sqrt{13}}{2}$

Now we need to calculate these values and use them to find the possible values of 'w'.

Understanding the Quadratic Formula

The quadratic formula is a fundamental tool in algebra. It allows us to solve for the roots of any quadratic equation, even when factoring isn't possible. In this case, it helps us determine the possible values of $\cos(w)$. Remember that the roots of the quadratic equation represent the points where the parabola crosses the x-axis. By applying the quadratic formula correctly, we find the potential values for the cosine of our angle. This is a critical step, and it's essential to make sure you're comfortable with the formula and how to apply it. Double-check your calculations to avoid any errors!

Step 3: Find the Values of w

Now we're getting to the heart of the problem: finding the values of 'w' that satisfy our equation. We have two potential values for $\cos(w)$ from the previous step:

1. $\cos(w) = \frac{-3 + \sqrt{13}}{2} \approx 0.30$
2. $\cos(w) = \frac{-3 - \sqrt{13}}{2} \approx -3.30$

Let's deal with these one at a time. For the first case, $\cos(w) \approx 0.30$. To find the angle 'w', we use the inverse cosine function (also known as arccosine), often denoted as $\cos^{-1}$ or arccos. Make sure your calculator is in radian mode! So,

$w = \cos^{-1}(0.30) \approx 1.27$ radians

However, remember that the cosine function is positive in both the first and fourth quadrants. So, there's another solution in the fourth quadrant. To find it, we use the fact that cosine is an even function and use the reference angle we found.

$w = 2\pi - 1.27 \approx 5.01$ radians

Now, let's consider the second case, $\cos(w) \approx -3.30$. Here's where we need to pause and think. The range of the cosine function is between -1 and 1, inclusive. Since -3.30 is outside of this range, there are no solutions for this case. Therefore, we only have two valid solutions.

Why we check the range

This is a crucial step to ensure our answers are valid. The range of the cosine function is [-1, 1]. This means that the cosine of any angle must fall between -1 and 1. Any value outside this range is impossible. By checking the range, we can quickly identify solutions that don't make sense and eliminate them. This prevents us from wasting time and effort on incorrect answers. Always remember to verify your results, especially when working with trigonometric functions, to avoid mathematical errors.

Step 4: State the Solutions

Alright, after all that work, we've found our solutions! We determined that the valid values of 'w' within the range $0 \leq w < 2\pi$ are:

$w \approx 1.27$ radians

$w \approx 5.01$ radians

So, the final answer, accurate to two decimal places, is: 1.27, 5.01.

Recap and Final Thoughts

Solving trigonometric equations can seem daunting, but by breaking it down into smaller steps, it becomes much more manageable. Here's a quick recap of what we did:

1. Used the Pythagorean Identity: We rewrote $\sin^2(w)$ in terms of $\cos^2(w)$ to create an equation with only one trigonometric function.
2. Solved the Quadratic Equation: We applied the quadratic formula to find the possible values of $\cos(w)$.
3. Found the Values of w: We used the inverse cosine function and considered all possible quadrants to find the angles 'w'. We also checked the range of the cosine function to ensure our solutions were valid.
4. Stated the Solutions: We presented our answers, rounded to two decimal places.

I hope this step-by-step guide has been helpful, guys. Remember to practice these types of problems to build your confidence and understanding. Keep an eye out for similar problems, and don't be afraid to break them down using the same principles. Happy solving!