Solving $10sin^2(x) - 3sin(x) - 1 = 0$: A Step-by-Step Guide

Hey guys! Let's break down how to solve this trigonometric equation: $10 ext{sin}^2(x) - 3 ext{sin}(x) - 1 = 0$. It looks a bit intimidating at first, but don't worry, we'll get through it together. We'll use a clever substitution to turn it into a quadratic equation, which we can then solve using factoring. So, grab your thinking caps, and let's dive in!

Understanding the Problem

Before we jump into the solution, let's make sure we understand what the problem is asking. We have a trigonometric equation involving the sine function, $\text{sin}(x)$. Our goal is to find the values of $x$ that satisfy the equation $10\text{sin}^2(x) - 3\text{sin}(x) - 1 = 0$. This means we're looking for angles $x$ where plugging $\text{sin}(x)$ into the equation makes it true.

The key to solving this equation lies in recognizing its similarity to a quadratic equation. We have a term with $\text{sin}^2(x)$, a term with $\text{sin}(x)$, and a constant term. This structure is very similar to the standard form of a quadratic equation, which is $ax^2 + bx + c = 0$. By making a simple substitution, we can transform our trigonometric equation into a quadratic equation that we can solve more easily. Remember, mathematics is all about spotting these patterns and using them to our advantage. The more you practice, the better you'll become at recognizing these connections and applying the appropriate techniques. So, let's keep this in mind as we proceed, and we'll see how this substitution simplifies things dramatically.

The Substitution Trick: Turning Trigonometry into Algebra

Okay, here’s the magic trick that simplifies everything. Let's substitute $u = \text{sin}(x)$. This means everywhere we see $\text{sin}(x)$ in the equation, we're going to replace it with the variable $u$. This might seem like a small step, but it makes a huge difference. By doing this, we're essentially turning a trigonometric equation into an algebraic equation, which is something we're probably more comfortable dealing with. So, let’s see what happens when we make this substitution.

When we substitute $u = \text{sin}(x)$ into our original equation, $10\text{sin}^2(x) - 3\text{sin}(x) - 1 = 0$, we get: $10u^2 - 3u - 1 = 0$. Look at that! Our trigonometric equation has transformed into a quadratic equation in terms of $u$. This is a quadratic equation in the standard form $au^2 + bu + c = 0$, where $a = 10$, $b = -3$, and $c = -1$. Now, we can use our knowledge of solving quadratic equations to find the values of $u$. This substitution technique is a common and powerful tool in mathematics. It allows us to simplify complex problems by transforming them into forms that we already know how to handle. The key is to identify the underlying structure of the equation and find a suitable substitution that makes the problem more manageable. Remember, this technique isn't limited to trigonometric equations; it can be applied in various areas of mathematics to simplify complex expressions and equations.

Solving the Quadratic Equation

Now that we have our quadratic equation, $10u^2 - 3u - 1 = 0$, we need to solve for $u$. There are a few ways we can do this, such as using the quadratic formula, but in this case, factoring is the most efficient method. Factoring involves breaking down the quadratic expression into the product of two binomials. This might sound intimidating if you haven't done it in a while, but with a little practice, it becomes a straightforward process. So, let's give it a try and see if we can factor this quadratic equation.

To factor the quadratic equation $10u^2 - 3u - 1 = 0$, we need to find two binomials that, when multiplied together, give us the original quadratic expression. This means we're looking for two expressions of the form $(Au + B)(Cu + D)$, where $A, B, C,$ and $D$ are constants. We need to find values for these constants such that $(Au + B)(Cu + D) = 10u^2 - 3u - 1$. A little trial and error, combined with a good understanding of factoring techniques, will help us find the correct binomials. After some careful consideration, we can factor the quadratic equation as follows: $(2u - 1)(5u + 1) = 0$. Notice how the product of the first terms in each binomial, $2u$ and $5u$, gives us $10u^2$. The product of the last terms, $-1$ and $1$, gives us $-1$. And the combination of the inner and outer products, $-5u$ and $2u$, gives us $-3u$. This confirms that our factoring is correct. Now that we have factored the equation, we can use the zero-product property to find the solutions for $u$.

Finding the Values of u

We've successfully factored the quadratic equation into $(2u - 1)(5u + 1) = 0$. Now, we can use the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. In other words, if $AB = 0$, then either $A = 0$ or $B = 0$ (or both). This property is a fundamental tool for solving equations that are expressed as a product of factors. It allows us to break down a single equation into multiple simpler equations, each of which we can solve independently. So, let’s apply this property to our factored equation and see what we get.

Applying the zero-product property to $(2u - 1)(5u + 1) = 0$, we get two separate equations:

1. $2u - 1 = 0$
2. $5u + 1 = 0$

Now, we can solve each of these equations for $u$. For the first equation, $2u - 1 = 0$, we add 1 to both sides to get $2u = 1$, and then divide both sides by 2 to get $u = \frac{1}{2}$. For the second equation, $5u + 1 = 0$, we subtract 1 from both sides to get $5u = -1$, and then divide both sides by 5 to get $u = -\frac{1}{5}$. So, we have found two possible values for $u$: $u = \frac{1}{2}$ and $u = -\frac{1}{5}$. But remember, we're not looking for the values of $u$; we're looking for the values of $x$. We need to reverse our substitution and go back to the original variable. This is a crucial step in solving any equation where we've made a substitution. We can't forget to undo the substitution and express our answers in terms of the original variables.

Reversing the Substitution: Back to Sine

We found the values of $u$, but remember that $u = \text{sin}(x)$. So, we need to substitute $\text{sin}(x)$ back in for $u$ to find the values of $x$. This is the crucial step of reversing the substitution. We initially made the substitution to simplify the equation, but now we need to go back to the original variable to answer the question that was asked. This process of substitution and back-substitution is a common technique in mathematics, allowing us to manipulate equations and solve them more easily. So, let’s see how it works in this case.

We have two values for $u$: $u = \frac{1}{2}$ and $u = -\frac{1}{5}$. Substituting $\text{sin}(x)$ back in, we get two new equations:

1. $\text{sin}(x) = \frac{1}{2}$
2. $\text{sin}(x) = -\frac{1}{5}$

Now we need to solve each of these equations for $x$. These are basic trigonometric equations that we can solve using our knowledge of the unit circle and the inverse sine function. We know that the sine function represents the y-coordinate of a point on the unit circle. So, we are looking for angles $x$ where the y-coordinate is equal to $\frac{1}{2}$ or $-\frac{1}{5}$. To find these angles, we'll use the inverse sine function, also known as arcsin. Remember, the inverse sine function gives us the angle whose sine is a given value. However, it's important to keep in mind that the sine function is periodic, meaning it repeats its values over and over again. Therefore, there will be infinitely many solutions to these equations. We'll need to consider the general solutions to account for all possible values of $x$.

Solving for x: Inverse Sine and General Solutions

Let's tackle the equations $\text{sin}(x) = \frac{1}{2}$ and $\text{sin}(x) = -\frac{1}{5}$ one by one. For $\text{sin}(x) = \frac{1}{2}$, we know from our knowledge of the unit circle that one solution is $x = \frac{\pi}{6}$ (which is 30 degrees). The sine function is positive in the first and second quadrants, so there's another solution in the second quadrant, which is $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$ (which is 150 degrees). These are the solutions within the interval $[0, 2\pi)$. However, since the sine function is periodic with a period of $2\pi$, we need to add multiples of $2\pi$ to these solutions to get the general solutions. Therefore, the general solutions for $\text{sin}(x) = \frac{1}{2}$ are: $x = \frac{\pi}{6} + 2\pi k$ and $x = \frac{5\pi}{6} + 2\pi k$, where $k$ is an integer.

Now, let's consider $\text{sin}(x) = -\frac{1}{5}$. Since $-\frac{1}{5}$ is not a standard value we encounter on the unit circle, we'll need to use the inverse sine function (arcsin) to find a solution. Using a calculator, we find that $\text{arcsin}(-\frac{1}{5}) \approx -0.2014$ radians. This is a solution in the fourth quadrant. The sine function is also negative in the third quadrant, so there's another solution in the third quadrant. To find this solution, we can use the fact that $\text{sin}(x) = \text{sin}(\pi - x)$. So, the other solution is $x = \pi - \text{arcsin}(-\frac{1}{5}) \approx \pi + 0.2014 \approx 3.3430$ radians. Again, we need to add multiples of $2\pi$ to these solutions to get the general solutions. Therefore, the general solutions for $\text{sin}(x) = -\frac{1}{5}$ are: $x \approx -0.2014 + 2\pi k$ and $x \approx 3.3430 + 2\pi k$, where $k$ is an integer. So, we've found the general solutions for both trigonometric equations, which give us all possible values of $x$ that satisfy the original equation.

Final Answer:

Alright, guys, we've made it to the end! We successfully solved the equation $10\text{sin}^2(x) - 3\text{sin}(x) - 1 = 0$. To recap, we used a clever substitution to transform the trigonometric equation into a quadratic equation. We then factored the quadratic equation, found the values of $u$, and reversed the substitution to get back to $\text{sin}(x)$. Finally, we solved for $x$ using the inverse sine function and considering the general solutions.

So, the solutions to the equation are:

• $x = \frac{\pi}{6} + 2\pi k$
• $x = \frac{5\pi}{6} + 2\pi k$
• $x \approx -0.2014 + 2\pi k$
• $x \approx 3.3430 + 2\pi k$

where $k$ is an integer. Phew! That was a bit of a journey, but we made it through together. Remember, the key to solving these types of problems is to break them down into smaller, more manageable steps. Don't be afraid to use substitutions, factoring, and other techniques to simplify the equation. And most importantly, practice, practice, practice! The more you solve these types of problems, the more comfortable you'll become with the process. You've got this!