Solving Trigonometric Equations: A Step-by-Step Guide

Hey guys! Today, we're diving into the fascinating world of trigonometry to tackle a problem that might seem daunting at first glance, but trust me, it's totally doable. We’re going to solve the trigonometric equation $-4 \cos^2 \theta + 9 \sin \theta + 3 = 3 \sin \theta - 3$ for all values of $0 \leq \theta < 2\pi$. Grab your calculators (or your mental math hats!), and let's get started!

Understanding the Trigonometric Equation

First off, let's break down this equation. Trigonometric equations can look a bit intimidating because they involve trigonometric functions like sine (sin), cosine (cos), tangent (tan), and their friends. But don't worry, the key is to simplify and use some fundamental trigonometric identities to make our lives easier. In this particular equation, we have both $\cos^2 \theta$ and $\sin \theta$, which means we need to find a way to express everything in terms of a single trigonometric function. This is where our handy trigonometric identities come in!

The equation we're dealing with is $-4 \cos^2 \theta + 9 \sin \theta + 3 = 3 \sin \theta - 3$. Our goal is to find all the values of $\theta$ (theta) that make this equation true within the interval $0 \leq \theta < 2\pi$. This interval represents one full revolution around the unit circle, so we're looking for solutions within this range. To solve this, we need to use the Pythagorean identity, which is a cornerstone in trigonometry. The Pythagorean identity states that $\sin^2 \theta + \cos^2 \theta = 1$. We can rearrange this identity to express $\cos^2 \theta$ in terms of $\sin^2 \theta$, which will help us simplify the equation.

Transforming the Equation Using Trigonometric Identities

Alright, let's put our math hats on and dive into the transformation process. Remember that Pythagorean identity we just talked about? $\sin^2 \theta + \cos^2 \theta = 1$. We can rearrange this to solve for $\cos^2 \theta$: $\cos^2 \theta = 1 - \sin^2 \theta$. This is gold because now we can substitute this expression into our original equation. So, let’s do it!

Our original equation is $-4 \cos^2 \theta + 9 \sin \theta + 3 = 3 \sin \theta - 3$. Replace $\cos^2 \theta$ with $(1 - \sin^2 \theta)$: $-4(1 - \sin^2 \theta) + 9 \sin \theta + 3 = 3 \sin \theta - 3$. Now, let's distribute the -4: $-4 + 4 \sin^2 \theta + 9 \sin \theta + 3 = 3 \sin \theta - 3$. Next, we'll move all terms to one side to set the equation to zero. This will make it look more like a quadratic equation, which we know how to solve. Subtract $3 \sin \theta$ from both sides and add 3 to both sides: $4 \sin^2 \theta + 9 \sin \theta - 3 \sin \theta - 4 + 3 + 3 = 0$. This simplifies to $4 \sin^2 \theta + 6 \sin \theta + 2 = 0$.

Simplifying to a Quadratic Form

Now, let's simplify this equation further. We've got a mix of terms, but it's shaping up nicely. We have $4 \sin^2 \theta + 6 \sin \theta + 2 = 0$. Notice anything familiar? This looks a lot like a quadratic equation, which is awesome because we have tools to solve those! To make it even clearer, let’s make a substitution. Let’s say $x = \sin \theta$. This transforms our equation into: $4x^2 + 6x + 2 = 0$.

See? Much friendlier now! This is a standard quadratic equation in the form $ax^2 + bx + c = 0$. To solve this, we can use factoring, the quadratic formula, or even completing the square. Factoring is often the quickest route if we can spot the factors easily. In this case, we can factor out a 2 from the entire equation, which simplifies things even further: $2(2x^2 + 3x + 1) = 0$. Dividing both sides by 2, we get $2x^2 + 3x + 1 = 0$.

Solving the Quadratic Equation

Okay, let's get our hands dirty and solve this quadratic equation. We have $2x^2 + 3x + 1 = 0$. We're going to try factoring first because it's often the fastest way. We need to find two numbers that multiply to 2 (the coefficient of $x^2$) and add up to 3 (the coefficient of $x$). Those numbers are 2 and 1. So, we can rewrite the middle term as $2x + x$:

$2x^2 + 2x + x + 1 = 0$. Now, we factor by grouping. Factor out $2x$ from the first two terms and 1 from the last two terms: $2x(x + 1) + 1(x + 1) = 0$. Notice that we have a common factor of $(x + 1)$, so we can factor that out: $(2x + 1)(x + 1) = 0$.

Great! Now we have the equation in factored form. To find the solutions, we set each factor equal to zero: $2x + 1 = 0$ or $x + 1 = 0$. Solving the first equation, $2x + 1 = 0$, we get $2x = -1$, so $x = -\frac{1}{2}$. Solving the second equation, $x + 1 = 0$, we get $x = -1$.

Finding the Values of Theta

Alright, we've solved for $x$, but remember, $x = \sin \theta$. So, we need to find the values of $\theta$ that satisfy $\sin \theta = -\frac{1}{2}$ and $\sin \theta = -1$ within the interval $0 \leq \theta < 2\pi$. This is where our knowledge of the unit circle comes in handy!

Let's start with $\sin \theta = -1$. Where on the unit circle is the sine (y-coordinate) equal to -1? That’s right, at $\theta = \frac{3\pi}{2}$. This is one of our solutions. Now, let's tackle $\sin \theta = -\frac{1}{2}$. The sine function is negative in the third and fourth quadrants. We know that $\sin \frac{\pi}{6} = \frac{1}{2}$, so the reference angle is $\frac{\pi}{6}$. In the third quadrant, the angle is $\pi + \frac{\pi}{6} = \frac{7\pi}{6}$. In the fourth quadrant, the angle is $2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$.

Final Solutions

Woohoo! We've made it to the finish line. We found all the values of $\theta$ that satisfy the original equation within the given interval. So, let’s gather our results and present them clearly. We found three solutions:

1. $\theta = \frac{3\pi}{2}$
2. $\theta = \frac{7\pi}{6}$
3. $\theta = \frac{11\pi}{6}$

So, there you have it! We successfully solved the trigonometric equation $-4 \cos^2 \theta + 9 \sin \theta + 3 = 3 \sin \theta - 3$ for $0 \leq \theta < 2\pi$. You guys did an amazing job following along. Remember, the key to solving these types of problems is to use trigonometric identities to simplify the equation, transform it into a quadratic form, solve the quadratic equation, and then use your knowledge of the unit circle to find the values of $\theta$. Keep practicing, and you'll become a trig equation-solving pro in no time!