Solving $6 Tan^3 Θ + 17 Tan^2 Θ + 4 Tan Θ - 12 = 0$

Introduction

In this comprehensive guide, we will delve into the process of solving the trigonometric equation $6 \tan ^3 \theta+17 \tan ^2 \theta+4 \tan \theta-12=0$ within the specified interval of $\frac{\pi}{2}<\theta<\pi$. This equation, a cubic polynomial in terms of $\tan \theta$, requires a strategic approach to find its solutions. Trigonometric equations like this often appear in various fields of mathematics, physics, and engineering, making it crucial to understand the methods to solve them. Our step-by-step explanation aims to provide a clear and concise understanding of the solution process, ensuring readers can confidently tackle similar problems. By the end of this guide, you will have a solid grasp of how to solve cubic trigonometric equations, enhancing your problem-solving skills in trigonometry. Whether you're a student, educator, or simply a math enthusiast, this guide offers valuable insights and techniques to expand your mathematical toolkit. Let's begin our journey by first understanding the complexities of the equation and then systematically working towards finding its roots. We will explore different methods, including factoring and using the quadratic formula, to arrive at the solutions. This detailed exploration will not only help solve the current problem but also build a strong foundation for solving other trigonometric equations. Stay tuned as we unravel the intricacies of this fascinating problem.

Understanding the Problem

To effectively solve the trigonometric equation $6 \tan ^3 \theta+17 \tan ^2 \theta+4 \tan \theta-12=0$ for $\frac{\pi}{2}<\theta<\pi$, we must first understand the problem thoroughly. This equation is a cubic polynomial in terms of $\tan \theta$, which means it involves the cube, square, and first power of the tangent function. The given interval, $\frac{\pi}{2}<\theta<\pi$, specifies that we are looking for solutions in the second quadrant of the unit circle. In the second quadrant, the tangent function is negative. This crucial piece of information helps us narrow down the possible solutions. The equation's complexity arises from the cubic term, which necessitates the use of algebraic techniques such as factoring or synthetic division to simplify it. Before diving into the algebraic manipulations, it's important to recognize that trigonometric equations often have multiple solutions due to the periodic nature of trigonometric functions. However, the specified interval limits our search to a specific range, making the problem more manageable. The strategy here is to treat $\tan \theta$ as a variable, say $x$, and solve the resulting cubic equation $6x^3 + 17x^2 + 4x - 12 = 0$. Once we find the roots of this polynomial, we can then determine the values of $\theta$ that satisfy the original equation within the given interval. This approach transforms a complex trigonometric problem into a more familiar algebraic one, allowing us to apply well-established methods for solving polynomial equations. By carefully analyzing the equation and the interval, we set the stage for a systematic solution process, ensuring we find all valid solutions.

Transforming the Equation

To make the equation $6 \tan ^3 \theta+17 \tan ^2 \theta+4 \tan \theta-12=0$ more manageable, we can perform a substitution. Let's substitute $x = \tan \theta$. This transforms our trigonometric equation into a standard cubic polynomial equation: $6x^3 + 17x^2 + 4x - 12 = 0$. This algebraic equation is now in a form that we can solve using various methods such as factoring, synthetic division, or numerical techniques. The advantage of this transformation is that it allows us to apply familiar algebraic tools to find the roots of the polynomial. Once we find the values of $x$, we can then substitute back $\tan \theta$ for $x$ and solve for $\theta$. The transformation simplifies the problem by separating the algebraic manipulation from the trigonometric context. This approach is common in solving complex trigonometric equations, where converting the equation into an algebraic form makes it easier to handle. By focusing on the polynomial equation, we can use techniques like the Rational Root Theorem to identify potential rational roots. The Rational Root Theorem helps us list possible rational solutions, which we can then test using synthetic division or direct substitution. This step is crucial because finding one root of the cubic equation allows us to reduce it to a quadratic equation, which is much easier to solve. Thus, transforming the equation is a pivotal step in our problem-solving strategy, setting the stage for finding the roots efficiently and accurately. Now that we have a polynomial equation, we can proceed with the algebraic techniques to find the values of $x$.

Solving the Cubic Equation

Now that we have transformed the trigonometric equation into a cubic equation, $6x^3 + 17x^2 + 4x - 12 = 0$, we need to find its roots. One effective method for solving cubic equations is to use the Rational Root Theorem. This theorem states that if a polynomial has a rational root $p/q$, then $p$ must be a factor of the constant term, and $q$ must be a factor of the leading coefficient. In our case, the constant term is -12, and the leading coefficient is 6. The factors of -12 are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$, and the factors of 6 are $\pm 1, \pm 2, \pm 3, \pm 6$. Therefore, the possible rational roots are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}, \pm \frac{1}{6}$. We can test these potential roots by substituting them into the equation or using synthetic division. Let's try $x = -2$. Substituting $x = -2$ into the equation, we get $6(-2)^3 + 17(-2)^2 + 4(-2) - 12 = 6(-8) + 17(4) - 8 - 12 = -48 + 68 - 8 - 12 = 0$. So, $x = -2$ is a root. Knowing this, we can perform synthetic division or factor out $(x + 2)$ from the cubic equation. Using synthetic division, we find that $6x^3 + 17x^2 + 4x - 12 = (x + 2)(6x^2 + 5x - 6)$. Now we have a quadratic equation, $6x^2 + 5x - 6 = 0$, which we can solve using the quadratic formula or factoring. This step is critical as it simplifies the problem significantly, allowing us to find the remaining roots with ease. By systematically applying the Rational Root Theorem and synthetic division, we have successfully reduced the cubic equation to a manageable quadratic equation.

Solving the Quadratic Equation and Finding tan θ

Having factored the cubic equation into $(x + 2)(6x^2 + 5x - 6) = 0$, we now need to solve the quadratic equation $6x^2 + 5x - 6 = 0$. We can solve this quadratic equation either by factoring or by using the quadratic formula. Let's try factoring first. We are looking for two numbers that multiply to $6 \times -6 = -36$ and add up to 5. These numbers are 9 and -4. So, we can rewrite the middle term as $5x = 9x - 4x$. The equation becomes $6x^2 + 9x - 4x - 6 = 0$. Now, we can factor by grouping: $3x(2x + 3) - 2(2x + 3) = 0$. This gives us $(3x - 2)(2x + 3) = 0$. Thus, the solutions for the quadratic equation are $x = \frac{2}{3}$ and $x = -\frac{3}{2}$. Now we have all the roots for the cubic equation: $x = -2, x = \frac{2}{3}$, and $x = -\frac{3}{2}$. Recall that we substituted $x = \tan \theta$. So, we have $\tan \theta = -2$, $\tan \theta = \frac{2}{3}$, and $\tan \theta = -\frac{3}{2}$. However, we are looking for solutions in the interval $\frac{\pi}{2} < \theta < \pi$, which is the second quadrant. In the second quadrant, the tangent function is negative. Therefore, we only consider the negative values of $\tan \theta$, which are $\tan \theta = -2$ and $\tan \theta = -\frac{3}{2}$. The positive value $\tan \theta = \frac{2}{3}$ is not in the specified interval. This step is crucial because it filters out extraneous solutions and focuses on the relevant roots within the given domain. By solving the quadratic equation and considering the quadrant in which we are looking for solutions, we have narrowed down the possible values of $\tan \theta$.

Finding θ in the Specified Interval

Now that we have the values of $\tan \theta$ as -2 and -\frac{3}{2}, we need to find the angles $\theta$ in the interval $\frac{\pi}{2} < \theta < \pi$ (the second quadrant). To find $\theta$, we use the inverse tangent function, also known as arctangent, denoted as $\arctan$ or $\tan^{-1}$. For $\tan \theta = -2$, we have $\theta = \arctan(-2)$. The arctangent function gives us the principal value, which lies in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$. Since -2 is negative, the principal value will be in the fourth quadrant. To find the corresponding angle in the second quadrant, we add $\pi$ to the principal value. Thus, $\theta_1 = \arctan(-2) + \pi$. Using a calculator, we find that $\arctan(-2) \approx -1.107$ radians. Adding $\pi$ to this value, we get $\theta_1 \approx -1.107 + \pi \approx 2.034$ radians. This value lies in the second quadrant, so it is a valid solution. Similarly, for $\tan \theta = -\frac{3}{2}$, we have $\theta = \arctan(-\frac{3}{2})$. The principal value of $\arctan(-\frac{3}{2})$ is also in the fourth quadrant. To find the corresponding angle in the second quadrant, we add $\pi$ to the principal value. Thus, $\theta_2 = \arctan(-\frac{3}{2}) + \pi$. Using a calculator, we find that $\arctan(-\frac{3}{2}) \approx -0.983$ radians. Adding $\pi$ to this value, we get $\theta_2 \approx -0.983 + \pi \approx 2.159$ radians. This value also lies in the second quadrant, so it is a valid solution. Therefore, the solutions to the equation $6 \tan ^3 \theta+17 \tan ^2 \theta+4 \tan \theta-12=0$ in the interval $\frac{\pi}{2} < \theta < \pi$ are approximately 2.034 radians and 2.159 radians. This step is crucial for obtaining the final answers, as it involves using the inverse trigonometric functions and considering the given interval to find the correct angles.

Conclusion

In conclusion, we have successfully solved the trigonometric equation $6 \tan ^3 \theta+17 \tan ^2 \theta+4 \tan \theta-12=0$ for $\frac{\pi}{2}<\theta<\pi$. Our approach involved transforming the trigonometric equation into a cubic polynomial equation by substituting $x = \tan \theta$. We then used the Rational Root Theorem to find one root of the cubic equation, which allowed us to factor it into a linear and a quadratic factor. Solving the quadratic equation yielded the remaining roots. We then substituted back $\tan \theta$ for $x$ and considered only the negative values since we were looking for solutions in the second quadrant. Finally, we used the inverse tangent function to find the angles $\theta$ in the specified interval. The solutions we found are approximately 2.034 radians and 2.159 radians. This problem demonstrates a powerful problem-solving strategy: transforming a complex equation into a simpler form that we can solve using familiar techniques. By combining algebraic manipulation with trigonometric concepts, we were able to systematically find the solutions. This approach is applicable to a wide range of trigonometric equations and highlights the importance of understanding both algebraic and trigonometric principles. Understanding these methods not only helps in solving specific problems but also enhances overall mathematical proficiency. We hope this step-by-step guide has provided you with a clear and concise understanding of how to solve cubic trigonometric equations. For further exploration of trigonometric identities and equation-solving techniques, you can visit reputable websites like Khan Academy Trigonometry.