Solving 2cos Θ + 1 = Sec Θ On [0°, 360°) A Step-by-Step Guide

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Solving trigonometric equations within a specific interval is a fundamental skill in trigonometry, with applications spanning various fields like physics, engineering, and computer graphics. In this comprehensive guide, we will delve into the process of solving the equation 2extcosθ+1=extsecθ2 ext{cos } θ + 1 = ext{sec } θ for solutions within the interval [0°,360°)[0°, 360°). We'll break down each step, providing clear explanations and justifications to ensure a thorough understanding. Trigonometric equations, unlike simple algebraic equations, involve trigonometric functions like sine, cosine, and tangent. These functions introduce a periodic nature to the equations, meaning they have multiple solutions within a given interval. This periodicity arises from the cyclical behavior of angles and their corresponding trigonometric values on the unit circle. To effectively solve trigonometric equations, it's crucial to leverage trigonometric identities and algebraic manipulation techniques. Trigonometric identities are equations that hold true for all values of the variables involved. These identities serve as powerful tools for simplifying complex trigonometric expressions and transforming equations into a more manageable form. By strategically applying these identities, we can often rewrite the equation in terms of a single trigonometric function, making it easier to isolate the variable and find the solutions. Additionally, algebraic manipulation techniques, such as factoring, substitution, and the quadratic formula, play a vital role in solving trigonometric equations. These techniques allow us to isolate the trigonometric function, solve for its value, and subsequently determine the angles that satisfy the equation. Understanding the unit circle is also paramount in solving trigonometric equations. The unit circle is a circle with a radius of 1 centered at the origin of the coordinate plane. It provides a visual representation of the trigonometric functions and their values for different angles. By referencing the unit circle, we can easily identify angles that correspond to specific trigonometric values, thus facilitating the solution process. In this article, we will explore the process of solving trigonometric equations with a particular emphasis on the equation 2extcosθ+1=extsecθ2 ext{cos } θ + 1 = ext{sec } θ. We will use a combination of trigonometric identities, algebraic manipulation techniques, and the unit circle to find all solutions within the interval [0°,360°)[0°, 360°). By mastering these techniques, you'll gain the ability to tackle a wide range of trigonometric equations and deepen your understanding of trigonometric functions and their applications.

Transforming the Equation Using Trigonometric Identities

In this section, we embark on the journey of solving the trigonometric equation 2extcosθ+1=extsecθ2 ext{cos } θ + 1 = ext{sec } θ within the interval [0°,360°)[0°, 360°). Our initial step involves transforming the equation into a more manageable form by leveraging trigonometric identities. The presence of both cosine ($ extcos } θ)andsecant() and secant ( ext{sec } θ$) functions in the equation complicates the direct solution process. To address this, we employ the reciprocal identity, which states that $ ext{sec } θ = rac{1}{ ext{cos } θ}$. This identity allows us to express the secant function in terms of cosine, thereby unifying the trigonometric functions present in the equation. Substituting rac{1}{ ext{cos } θ} for $ ext{sec } θ$ in the original equation, we obtain $2 ext{cos θ + 1 = rac1}{ ext{cos } θ}$. Now, the equation solely involves the cosine function, which simplifies our task significantly. To further streamline the equation, we aim to eliminate the fraction. We achieve this by multiplying both sides of the equation by $ ext{cos } θ$. This operation yields $(2 ext{cos θ + 1) extcos } θ = rac{1}{ ext{cos } θ} ext{cos } θ$. Simplifying the equation, we get $2 ext{cos^2 θ + extcos } θ = 1$. This transformation results in a quadratic equation in terms of $ ext{cos } θ$. A quadratic equation is a polynomial equation of degree two, typically expressed in the form ax2+bx+c=0ax^2 + bx + c = 0, where a, b, and c are constants. Recognizing the quadratic form of our equation is crucial because it allows us to apply standard algebraic techniques for solving quadratic equations. To prepare the equation for solving, we rearrange it by subtracting 1 from both sides $2 ext{cos^2 θ + extcos } θ - 1 = 0$. This rearrangement sets the stage for the next step factoring the quadratic expression. Factoring is the process of expressing a polynomial as a product of simpler polynomials. In this case, we seek two binomials whose product equals $2 ext{cos^2 θ + ext{cos } θ - 1$. Factoring the quadratic expression is a key step in solving the equation. By expressing the quadratic as a product of two factors, we can utilize 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. This property allows us to break down the quadratic equation into two simpler linear equations, which are much easier to solve. In the subsequent section, we will delve into the process of factoring the quadratic expression and applying the zero-product property to find the solutions for $ ext{cos } θ$.

Factoring and Solving the Quadratic Equation

Having transformed the original equation into a quadratic form, 2extcos2θ+extcosθ−1=02 ext{cos}^2 θ + ext{cos } θ - 1 = 0, we now focus on solving this equation to find the values of $ extcos } θ$ that satisfy it. The key to solving this quadratic equation lies in factoring the expression on the left-hand side. Factoring involves expressing the quadratic expression as a product of two binomials. In this case, we seek two binomials that, when multiplied together, yield 2extcos2θ+extcosθ−12 ext{cos}^2 θ + ext{cos } θ - 1. After careful consideration, we find that the quadratic expression can be factored as follows $(2 ext{cos θ - 1)( extcos } θ + 1) = 0$. This factorization is a crucial step because it allows us to apply the zero-product property. The zero-product property 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=0AB = 0, then either A=0A = 0 or B=0B = 0 (or both). Applying this property to our factored equation, we set each factor equal to zero $2 ext{cos θ - 1 = 0$ or $ extcos } θ + 1 = 0$. This step transforms our single quadratic equation into two simpler linear equations, which are much easier to solve. Let's first consider the equation 2extcosθ−1=02 ext{cos } θ - 1 = 0. To isolate $ ext{cos } θ$, we add 1 to both sides $2 ext{cos θ = 1$. Then, we divide both sides by 2: $ extcos } θ = rac{1}{2}$. This equation tells us that we need to find the angles θθ within the interval [0°,360°)[0°, 360°) for which the cosine function equals rac{1}{2}. Now, let's turn our attention to the second equation, $ ext{cos } θ + 1 = 0$. To isolate $ ext{cos } θ$, we subtract 1 from both sides $ ext{cos θ = -1$. This equation indicates that we need to find the angles θθ within the interval [0°,360°)[0°, 360°) for which the cosine function equals -1. In the next section, we will utilize the unit circle to identify the angles that satisfy these two equations, thereby finding the solutions for θθ.

Utilizing the Unit Circle to Determine Solutions

With the values of $ extcos } θ$ determined from the factored equation, $ ext{cos } θ = rac{1}{2}$ and $ ext{cos } θ = -1$, we now employ the unit circle to identify the angles θθ within the interval [0°,360°)[0°, 360°) that correspond to these cosine values. The unit circle is a circle with a radius of 1 centered at the origin of the coordinate plane. It provides a visual representation of the trigonometric functions and their values for different angles. For any point on the unit circle, the x-coordinate represents the cosine of the angle, and the y-coordinate represents the sine of the angle. Let's first consider the equation $ ext{cos } θ = rac{1}{2}$. We need to find the angles θθ in the interval [0°,360°)[0°, 360°) where the x-coordinate on the unit circle is rac{1}{2}. Recall that cosine is positive in the first and fourth quadrants. By examining the unit circle, we find two angles that satisfy this condition θ=60°θ = 60° and θ=300°θ = 300°. These angles correspond to the points on the unit circle where the x-coordinate is $ rac{12}$. Next, we consider the equation $ ext{cos } θ = -1$. We need to find the angles θθ in the interval [0°,360°)[0°, 360°) where the x-coordinate on the unit circle is -1. Examining the unit circle, we find only one angle that satisfies this condition θ=180°θ = 180°. This angle corresponds to the point on the unit circle where the x-coordinate is -1. Therefore, the solutions to the equation $2 ext{cos θ + 1 = ext{sec } θ$ within the interval [0°,360°)[0°, 360°) are θ=60°θ = 60°, θ=180°θ = 180°, and θ=300°θ = 300°. These angles represent the values of θθ that, when substituted into the original equation, make the equation true. In summary, by transforming the equation using trigonometric identities, factoring the resulting quadratic equation, and utilizing the unit circle, we have successfully found all the solutions to the equation 2extcosθ+1=extsecθ2 ext{cos } θ + 1 = ext{sec } θ within the specified interval. This process demonstrates the power of combining trigonometric identities, algebraic techniques, and the visual representation of the unit circle to solve trigonometric equations. This comprehensive approach provides a solid foundation for tackling a wide range of trigonometric problems.

Expressing the Solution Set

Having identified the solutions to the equation 2extcosθ+1=extsecθ2 ext{cos } θ + 1 = ext{sec } θ within the interval [0°,360°)[0°, 360°), which are θ=60°θ = 60°, θ=180°θ = 180°, and θ=300°θ = 300°, we now express these solutions as a solution set. The solution set is a set containing all the values that satisfy the given equation. In this case, the solution set consists of the angles 60°60°, 180°180°, and 300°300°. We represent the solution set using set notation, which involves enclosing the elements of the set within curly braces. Therefore, the solution set for the equation 2extcosθ+1=extsecθ2 ext{cos } θ + 1 = ext{sec } θ in the interval [0°,360°)[0°, 360°) is: {60°, 180°, 300°}. This set notation provides a concise and unambiguous way to represent the complete solution to the equation. It clearly indicates all the angles within the specified interval that satisfy the given trigonometric equation. The solution set is an important concept in mathematics, particularly in the context of solving equations. It provides a complete and organized representation of all the possible solutions, allowing for a clear understanding of the equation's behavior and its solutions. In this case, the solution set {60°, 180°, 300°} represents all the angles within the interval [0°,360°)[0°, 360°) that satisfy the equation 2extcosθ+1=extsecθ2 ext{cos } θ + 1 = ext{sec } θ. This completes the process of solving the trigonometric equation and expressing the solution set. By following this systematic approach, you can effectively solve a wide range of trigonometric equations and accurately represent their solutions using set notation.

Conclusion

In conclusion, we have successfully solved the trigonometric equation 2extcosθ+1=extsecθ2 ext{cos } θ + 1 = ext{sec } θ for solutions over the interval [0°,360°)[0°, 360°). Our approach involved a series of key steps, each building upon the previous one to arrive at the final solution. We began by transforming the equation using the reciprocal identity, expressing secant in terms of cosine. This allowed us to consolidate the trigonometric functions into a single term, simplifying the equation. Next, we rearranged the equation into a quadratic form, which is a familiar structure that can be solved using standard algebraic techniques. We then factored the quadratic expression, which enabled us to apply the zero-product property. This property allowed us to break down the quadratic equation into two simpler linear equations, making the solution process more manageable. Subsequently, we solved each linear equation to find the values of $ extcos } θ$. With these values in hand, we turned to the unit circle, a powerful visual tool for understanding trigonometric functions. By examining the unit circle, we identified the angles within the specified interval that corresponded to the calculated cosine values. This step yielded the solutions to the equation θ=60°θ = 60°, θ=180°θ = 180°, and θ=300°θ = 300°. Finally, we expressed these solutions as a solution set, {60°, 180°, 300°, providing a concise and comprehensive representation of all the solutions. This entire process highlights the importance of a multifaceted approach to solving trigonometric equations. It demonstrates the power of combining trigonometric identities, algebraic manipulation, and the visual aid of the unit circle. By mastering these techniques, you can confidently tackle a wide range of trigonometric problems and gain a deeper understanding of trigonometric functions and their applications. The ability to solve trigonometric equations is a valuable skill in various fields, including mathematics, physics, engineering, and computer science. It allows us to model and analyze periodic phenomena, such as oscillations, waves, and cycles. The techniques presented in this guide provide a solid foundation for further exploration of trigonometric concepts and their applications in real-world scenarios. By continuing to practice and apply these methods, you can enhance your problem-solving skills and develop a strong command of trigonometry.