Solving Sin(x)(sin(x)-1) = 0 A Step-by-Step Guide

Solving trigonometric equations is a fundamental skill in mathematics, especially in fields like calculus, physics, and engineering. One such equation is $\sin(x)(\sin(x) - 1) = 0$. This equation may seem simple at first glance, but a thorough understanding of trigonometric functions and their properties is essential to arrive at the correct solutions. This article provides a detailed, step-by-step solution to this equation, ensuring that the underlying principles are clear and accessible. By breaking down the equation and applying relevant trigonometric identities and concepts, we aim to offer a comprehensive guide for students and enthusiasts alike. This guide will not only solve the given equation but also equip readers with the tools to tackle similar problems with confidence. Understanding trigonometric equations is crucial as they appear in numerous contexts, ranging from modeling periodic phenomena to solving geometric problems. Thus, mastering the techniques to solve these equations is an invaluable asset for anyone studying mathematics or related disciplines.

The given equation, $\sin(x)(\sin(x) - 1) = 0$, is a trigonometric equation that involves the sine function. The primary goal is to find all values of $x$ that satisfy this equation. This involves understanding the properties of the sine function, its range, and its periodicity. We will also explore the unit circle and how it relates to the sine function, providing a visual aid to comprehending the solutions. By systematically examining the equation, we will identify the conditions under which it holds true and determine the general solutions that encompass all possible values of $x$. This process will illustrate the importance of considering the periodic nature of trigonometric functions and how to express solutions in a general form using integer multiples of the period. This thorough approach ensures that no solutions are missed and that the mathematical reasoning is sound and transparent.

The equation $\sin(x)(\sin(x) - 1) = 0$ can be solved by applying 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 this case, the two factors are $\sin(x)$ and $(\sin(x) - 1)$. Therefore, the equation is satisfied if either $\sin(x) = 0$ or $\sin(x) - 1 = 0$. We will examine each of these cases separately to find the solutions for $x$. This approach simplifies the problem by breaking it down into smaller, more manageable parts. The first case, $\sin(x) = 0$, corresponds to the values of $x$ where the sine function is zero. The second case, $\sin(x) - 1 = 0$, simplifies to $\sin(x) = 1$, which corresponds to the values of $x$ where the sine function is equal to one. By solving these two simpler equations, we can combine the results to obtain the complete set of solutions for the original equation. This method demonstrates a common and effective strategy for solving algebraic and trigonometric equations, highlighting the importance of recognizing and utilizing fundamental properties and techniques.

Step-by-Step Solution

Case 1: sin(x) = 0

The first case to consider is when $\sin(x) = 0$. We need to find all values of $x$ for which the sine function equals zero. Recall that the sine function represents the y-coordinate of a point on the unit circle. Therefore, $\sin(x) = 0$ when the point lies on the x-axis. This occurs at angles that are integer multiples of $\pi$. Specifically, the sine function is zero at $x = 0$, $x = \pi$, $x = 2\pi$, and so on, as well as at $x = -\pi$, $x = -2\pi$, and so on. To express this general solution mathematically, we use the form $x = n\pi$, where $n$ is an integer. This notation captures all the angles at which the sine function is zero. Understanding this concept is crucial for solving trigonometric equations, as it highlights the periodic nature of the sine function and the importance of representing solutions in a general form that accounts for all possible values.

To find the solutions for $x$ when $\sin(x) = 0$, it's helpful to visualize the unit circle. The sine function corresponds to the y-coordinate of a point on the unit circle. The y-coordinate is zero at the points where the unit circle intersects the x-axis. These intersections occur at angles of $0$ radians, $\pi$ radians, $2\pi$ radians, and so on, as well as at $-\pi$ radians, $-2\pi$ radians, and so on. These angles are all integer multiples of $\pi$. Therefore, the general solution for $\sin(x) = 0$ can be expressed as $x = n\pi$, where $n$ is an integer. This means that any angle of the form $n\pi$, where $n$ is an integer, will satisfy the equation $\sin(x) = 0$. This understanding is fundamental for solving trigonometric equations and is a cornerstone of trigonometric analysis.

Therefore, the general solution for $\sin(x) = 0$ is given by:

$x = n\pi$, where $n$ is an integer.

This solution encompasses all angles for which the sine function is zero, taking into account the periodic nature of the sine function. The integer $n$ represents the number of full or half rotations around the unit circle, ensuring that we capture all possible solutions. This general form is crucial for accurately representing the solutions to trigonometric equations, as it accounts for the infinite number of angles that satisfy the condition $\sin(x) = 0$. This approach demonstrates the power of using general solutions to fully describe the set of all possible values that satisfy a trigonometric equation, reinforcing the importance of understanding the underlying principles of trigonometric functions and their behavior.

Case 2: sin(x) - 1 = 0

Now, let's consider the second case: $\sin(x) - 1 = 0$. By adding 1 to both sides of the equation, we get $\sin(x) = 1$. This means we are looking for all values of $x$ where the sine function equals 1. Recall that the sine function represents the y-coordinate of a point on the unit circle. The y-coordinate is 1 at the point (0, 1) on the unit circle, which corresponds to an angle of $\frac{\pi}{2}$ radians. However, due to the periodic nature of the sine function, there are infinitely many angles that have the same sine value. The sine function repeats every $2\pi$ radians, so we need to consider all angles that are coterminal with $\frac{\pi}{2}$. These angles can be represented as $\frac{\pi}{2} + 2n\pi$, where $n$ is an integer.

To find the solutions for $x$ when $\sin(x) = 1$, we need to identify the angles on the unit circle where the y-coordinate is equal to 1. This occurs at the point (0, 1), which corresponds to an angle of $\frac{\pi}{2}$ radians. Because the sine function is periodic with a period of $2\pi$, we need to account for all angles that are coterminal with $\frac{\pi}{2}$. These angles can be represented by adding integer multiples of $2\pi$ to $\frac{\pi}{2}$. Thus, the general solution for $\sin(x) = 1$ is given by $x = \frac{\pi}{2} + 2n\pi$, where $n$ is an integer. This solution captures all the angles at which the sine function equals 1, considering the periodic nature of the function and the infinite number of angles that satisfy the condition. This systematic approach ensures that all possible solutions are accounted for, reinforcing the importance of understanding periodicity in solving trigonometric equations.

Therefore, the general solution for $\sin(x) = 1$ is given by:

$x = \frac{\pi}{2} + 2n\pi$, where $n$ is an integer.

This solution represents all angles where the sine function equals 1, taking into account the periodic nature of the sine function. The term $2n\pi$ accounts for all full rotations around the unit circle, ensuring that we capture all possible solutions. This comprehensive approach highlights the importance of considering periodicity when solving trigonometric equations and demonstrates how to express solutions in a general form that includes all possible values. By understanding and applying this method, students can confidently solve similar equations and gain a deeper appreciation for the behavior of trigonometric functions.

Combining the Solutions

To find the complete solution to the original equation $\sin(x)(\sin(x) - 1) = 0$, we need to combine the solutions from both cases. From Case 1, we found that $\sin(x) = 0$ when $x = n\pi$, where $n$ is an integer. From Case 2, we found that $\sin(x) = 1$ when $x = \frac{\pi}{2} + 2n\pi$, where $n$ is an integer. These two sets of solutions together give us all the values of $x$ that satisfy the original equation. The solutions from Case 1 represent the angles where the sine function is zero, while the solutions from Case 2 represent the angles where the sine function is one. By combining these solutions, we ensure that we have accounted for all possibilities and have a complete set of answers.

The solutions from Case 1, $x = n\pi$, represent the angles at which the sine function is zero. These angles correspond to the points where the unit circle intersects the x-axis. The solutions from Case 2, $x = \frac{\pi}{2} + 2n\pi$, represent the angles at which the sine function is one. This angle corresponds to the point (0, 1) on the unit circle. Combining these solutions provides a comprehensive set of angles that satisfy the original equation. It is crucial to include both sets of solutions to ensure that no possible values of $x$ are missed. This systematic approach of breaking down the equation into cases and then combining the solutions is a powerful technique for solving trigonometric equations and demonstrates a thorough understanding of the properties of trigonometric functions.

Therefore, the complete solution to the equation $\sin(x)(\sin(x) - 1) = 0$ is the union of the solutions from both cases:

$x = n\pi$ and $x = \frac{\pi}{2} + 2n\pi$, where $n$ is an integer.

This comprehensive solution set includes all possible values of $x$ that satisfy the given equation. It is essential to express the solutions in this general form to account for the periodic nature of the sine function and to ensure that all possible angles are included. This final answer demonstrates a thorough understanding of the problem and the methods used to solve it, reinforcing the importance of considering all possible cases and combining solutions to arrive at a complete and accurate answer.

Conclusion

In conclusion, to solve the trigonometric equation $\sin(x)(\sin(x) - 1) = 0$, we applied the zero-product property and considered two separate cases. The first case, $\sin(x) = 0$, yielded the general solution $x = n\pi$, where $n$ is an integer. The second case, $\sin(x) - 1 = 0$, led to the general solution $x = \frac{\pi}{2} + 2n\pi$, where $n$ is an integer. By combining these solutions, we obtained the complete set of solutions for the original equation. This process demonstrates a systematic approach to solving trigonometric equations, highlighting the importance of understanding the properties of trigonometric functions and their periodic nature. The ability to solve such equations is fundamental in many areas of mathematics and its applications, making this a crucial skill for students and professionals alike.

This step-by-step guide provides a clear and comprehensive method for solving the equation $\sin(x)(\sin(x) - 1) = 0$, ensuring that all possible solutions are accounted for. The breakdown into cases and the use of general solutions emphasize the importance of understanding the periodic nature of trigonometric functions. This method can be applied to a wide range of trigonometric equations, making it a valuable tool for anyone studying or working with mathematics. The detailed explanation and step-by-step approach aim to make the solution accessible to a broad audience, fostering a deeper understanding of trigonometric concepts and problem-solving techniques.

Therefore, the correct answer to the equation $\sin(x)(\sin(x) - 1) = 0$ is:

D. $x = n\pi$, $x = \frac{\pi}{2} + 2n\pi$, where $n$ is an integer.

This solution encapsulates all possible values of $x$ that satisfy the given equation, demonstrating a complete and accurate understanding of the problem and its solution. The methodical approach used in this article provides a solid foundation for tackling more complex trigonometric equations and reinforces the importance of clear and systematic problem-solving techniques. By mastering these methods, individuals can confidently approach trigonometric problems and apply their knowledge to various mathematical and real-world contexts.