Solving Trigonometric Equations: A Step-by-Step Guide

by Andrew McMorgan 54 views

Hey Plastik Magazine readers! Let's dive into some math today, specifically solving trigonometric equations. Don't worry, it's not as scary as it sounds. We'll break down a problem step-by-step to make it super easy to understand. We're going to tackle a question that involves a quadratic equation with a trigonometric function, which is a very common type of problem in algebra and trigonometry. Ready? Let's get started!

Understanding the Problem: The Foundation of Trigonometric Equations

Trigonometric equations are equations that involve trigonometric functions, like sine, cosine, tangent, etc. Solving them means finding the values of the variable (usually an angle, often represented by x) that make the equation true. The equation we're looking at is: $2 \sin ^2(x)-5 \sin (x)-3=0$. This might look a little intimidating at first glance, but trust me, we can handle it! The key here is recognizing the structure of the equation. Notice how we have a sin⁡2(x)\sin^2(x), a sin⁡(x)\sin(x), and a constant term? This is a classic quadratic equation in disguise. That's why we're going to use a clever trick to make it easier to solve. We're going to use substitution to transform it into a more familiar form.

The core concept revolves around the substitution of $u=\sin(x)$. By doing this, we're simplifying the appearance of the equation, making it easier to manipulate and solve. This is a common technique used in math to simplify complex expressions and equations. So, the first step is always about understanding what the equation is asking us to do, and what tools we can use to simplify the process. This specific type of problem is designed to test your knowledge of quadratic equations and trigonometric functions, and your ability to apply substitution as a problem-solving strategy. Always make sure you understand the fundamental concepts and the relationships between the elements of the equation before starting to solve, and then you'll find that it's much easier to approach.

The Role of Substitution in Trigonometric Equations

Substitution is a powerful technique in algebra and calculus. It involves replacing a part of an equation with a new variable to simplify the equation. In our case, we're substituting $u$ for $\sin(x)$. This transforms our equation from something that might initially look complex into something we can solve more easily. By using substitution, we are not changing the equation, but only changing its appearance to make it more manageable. Imagine it as putting on a different set of clothes. The person remains the same, but the outfit is different. The same goes for the equation: it remains equivalent, but looks simpler. The main advantage of substitution is that it allows us to convert a problem we don't know how to solve into one that we are familiar with. This approach is not only applicable to trigonometry but can also be widely used in other fields of mathematics. Understanding the role of substitution is critical to being able to effectively solve this equation and similar equations in the future.

The Substitution Step: Unveiling the Quadratic Equation

Let's put the substitution into action. Given that $u=\sin(x)$, we can replace every instance of $\sin(x)$ in the original equation with $u$. This gives us:

2u2−5u−3=02u^2 - 5u - 3 = 0

See how much cleaner that looks? Now we have a standard quadratic equation in terms of $u$. The next step is to solve this quadratic equation. You could use several methods to do this, such as factoring, completing the square, or using the quadratic formula. But in our case, the easiest method is factoring. Factoring involves finding two expressions that, when multiplied together, give us the original quadratic expression. The whole purpose here is to rewrite the original equation in an equivalent form that is easier to work with. So, now, we have the simplified equation, and we can solve it in a more straightforward manner.

Transforming the Equation with Substitution

Once we've made the substitution, the original trigonometric equation is transformed. Now, the main goal is to solve the newly formed quadratic equation for the variable $u$. Solving the quadratic equation means finding the values of $u$ that satisfy the equation. Because this equation has a standard quadratic form, we have the option of factoring, completing the square, or using the quadratic formula. Since the equation is relatively simple, factoring is generally the easiest option. You will notice that the key element is to recognize that we are working with an equation that is quadratic in form, making it possible to apply familiar methods to solve it. Once we solve for $u$, we will be able to determine the corresponding values of $\sin(x)$, and thus, solve for $x$.

Factoring the Quadratic Equation: Finding the Right Combination

Now, let's factor the quadratic equation $2u^2 - 5u - 3 = 0$. We're looking for two binomials that multiply to give us this equation. After a little bit of trial and error (or by using factoring techniques), we find that:

(2u+1)(u−3)=0(2u + 1)(u - 3) = 0

To check if you factored correctly, you can multiply these two binomials back together to make sure that they result in the original quadratic equation. If the answer is the same, then that means the factorization is correct. This is what we call verification, and it is a good habit to ensure we do not make mistakes. Now we have two factors, we can proceed to the next step, which is to solve each factor for $u$.

Understanding the Factoring Process

The ability to factor quadratic expressions is a fundamental skill in algebra. The process involves breaking down the quadratic expression into the product of two binomials. Each binomial will contain the variable $u$, along with some constant values. By factoring, we're essentially rewriting the original expression in a different form. It is the key step that allows us to find the values of $u$ that satisfy the equation. Once the equation is factored, we can set each factor equal to zero and solve for $u$. If the original equation could not have been factored, we would have had to use other methods, such as the quadratic formula. The factoring process depends heavily on your skill in algebra and is an important skill to master to become fluent in mathematics.

Solving for u: Uncovering the Solutions

Now that we've factored the equation, we can find the values of $u$. Set each factor equal to zero and solve for $u$:

  1. 2u + 1 = 0$ => $u = -1/2

  2. u - 3 = 0$ => $u = 3

So, we have two possible values for $u$: $-1/2$ and $3$. But remember, $u = \sin(x)$. Now, we need to think about what these values mean in terms of the original trigonometric equation.

The Significance of the Solutions for u

Once the quadratic equation is factored and solved for $u$, we have found the values of the sine function that satisfy the original equation. We should remember that $u$ represents $\sin(x)$. Therefore, the solutions for $u$ give us the possible values for $\sin(x)$. In our case, we have two possible values: $-1/2$ and $3$. However, we must remember the range of the sine function. The sine function, $\sin(x)$, has a range between $-1$ and $1$. Therefore, any solution for $u$ that falls outside of this range is not valid. The values we obtained for $u$ give us an opportunity to verify the validity of our solutions and to assess whether we need to eliminate any of them. The purpose of this step is to find potential solutions for the variable in the original trigonometric equation.

Connecting Back to x: The Final Step

We know that $u = \sin(x)$. Our solutions for $u$ are $-1/2$ and $3$. Since the sine function can only output values between -1 and 1, the solution $u = 3$ is not possible. So, we're left with $\sin(x) = -1/2$. To find the values of $x$ that satisfy this, we need to consider the unit circle or use the inverse sine function. The values of $x$ that satisfy this equation are the angles where the sine function equals -1/2. Because of the periodicity of the trigonometric functions, there will be multiple solutions. This step requires knowledge of the unit circle or the inverse sine function. The point of this is to link the solutions of $u$ to the variable of the original equation, completing the solution of the problem.

The Final Stage of Solving the Equation

Once we have solved the quadratic equation for $u$, we are ready to find the values of $x$ that satisfy the original trigonometric equation. We know that $u = \sin(x)$. Therefore, we need to find the angles $x$ for which $\sin(x)$ equals the solutions we found for $u$. In our case, one solution for $u$ was $-1/2$. This means we need to find the values of $x$ for which $\sin(x) = -1/2$. We can find those values by using the unit circle or the inverse sine function. The solutions are angles located in the third and fourth quadrants of the unit circle, where the sine function is negative. Remember that, due to the periodic nature of the sine function, there are infinitely many solutions. This involves a critical review of the range and the domain of the sine function. This final step is crucial to correctly solve the trigonometric equation and to find the values of the variable $x$ that make the original equation true.

The Answer

Looking back at our original problem, we were asked which of the following is equivalent to the given equation: $2 \sin ^2(x)-5 \sin (x)-3=0$ when $u=\sin (x)$. Based on our work above, the factored form of the equation is $(2u + 1)(u - 3) = 0$. Therefore, the correct answer is A. $(2 u+1)(u-3)=0$.

I hope that clears things up, guys! Keep practicing, and you'll become a pro at solving trigonometric equations. Peace out!