Solving Trigonometric Equations A Step-by-Step Guide To \tan^2 Θ + 7 \tan Θ + 9 = 0

In the realm of mathematics, particularly trigonometry, solving equations is a fundamental skill. Trigonometric equations, which involve trigonometric functions such as sine, cosine, and tangent, often arise in various scientific and engineering applications. This article delves into the process of solving a specific trigonometric equation, $\tan^2 θ + 7 \tan θ + 9 = 0$, over the interval $[0^{\circ}, 360^{\circ})$. We will explore the underlying concepts, step-by-step solutions, and provide a comprehensive understanding of how to tackle such problems. Before diving into the specifics, let's establish a strong foundation by revisiting key trigonometric concepts and techniques.

Trigonometric Equations: A Brief Overview

Trigonometric equations are equations that involve trigonometric functions of an unknown angle. Solving these equations means finding the values of the angle that satisfy the equation. Unlike algebraic equations, trigonometric equations often have infinitely many solutions due to the periodic nature of trigonometric functions. However, when we restrict the interval, such as $[0^{\circ}, 360^{\circ})$, we limit the solutions to a specific range, making the problem more manageable. Solving trigonometric equations is not just a mathematical exercise; it has practical implications in various fields. For instance, in physics, these equations are used to model oscillatory motions, such as the swing of a pendulum or the vibration of a string. In engineering, they are crucial for designing and analyzing systems involving periodic phenomena, like alternating current circuits or mechanical vibrations. Mastering the art of solving trigonometric equations opens doors to understanding and solving real-world problems.

Key Trigonometric Concepts

Before we tackle the equation $\tan^2 θ + 7 \tan θ + 9 = 0$, let's refresh some essential trigonometric concepts:

1. Tangent Function: The tangent function, denoted as $\tan θ$, is defined as the ratio of the sine of the angle to the cosine of the angle: $\tan θ = \frac{\sin θ}{\cos θ}$. It represents the slope of the line that passes through the origin and the point on the unit circle corresponding to the angle θ. The tangent function has a period of $180^{\circ}$ or π radians, meaning its values repeat every $180^{\circ}$.

2. Unit Circle: The unit circle is a circle with a radius of 1 centered at the origin in the Cartesian plane. It is a powerful tool for understanding trigonometric functions because the coordinates of any point on the unit circle can be expressed in terms of sine and cosine. Specifically, if a point on the unit circle corresponds to an angle θ, then its coordinates are $(\cos θ, \sin θ)$. The unit circle provides a visual representation of the values of sine and cosine for all angles, making it easier to solve trigonometric equations.

3. Quadratic Equations: A quadratic equation is a polynomial equation of the second degree, generally expressed in the form $ax^2 + bx + c = 0$, where a, b, and c are constants, and a ≠ 0. Quadratic equations have at most two solutions, which can be real or complex numbers. Solving quadratic equations is a fundamental skill in algebra, and it plays a crucial role in solving many trigonometric equations. We can use methods like factoring, completing the square, or the quadratic formula to find the solutions of a quadratic equation.

4. Quadratic Formula: The quadratic formula is a method for solving quadratic equations of the form $ax^2 + bx + c = 0$. The formula is given by:

   $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

   This formula provides the two solutions for x, which may be real or complex, depending on the value of the discriminant ($b^2 - 4ac$). If the discriminant is positive, there are two distinct real solutions. If it is zero, there is exactly one real solution. If it is negative, there are two complex solutions. The quadratic formula is a versatile tool that can be used to solve any quadratic equation, regardless of whether it can be factored or not.

5. Inverse Tangent Function: The inverse tangent function, denoted as $\arctan$ or $\tan^{-1}$, is the inverse function of the tangent function. It takes a value as input and returns the angle whose tangent is that value. The range of the inverse tangent function is typically restricted to the interval $(-90^{\circ}, 90^{\circ})$ or $(-\frac{π}{2}, \frac{π}{2})$ to ensure that it is a function (i.e., each input has a unique output). The inverse tangent function is essential for finding the angles that satisfy a trigonometric equation involving the tangent function.

Step-by-Step Solution of $\tan^2 θ + 7 \tan θ + 9 = 0$

Now that we have the necessary background, let's dive into solving the equation $\tan^2 θ + 7 \tan θ + 9 = 0$ over the interval $[0^{\circ}, 360^{\circ})$.

Step 1: Recognize the Quadratic Form

The given equation, $\tan^2 θ + 7 \tan θ + 9 = 0$, is a quadratic equation in terms of $\tan θ$. To make this more apparent, we can substitute $x = \tan θ$. This substitution transforms the equation into a familiar quadratic form:

$$x^2 + 7x + 9 = 0$$

Recognizing the quadratic form is a crucial first step because it allows us to apply the well-established methods for solving quadratic equations.

Step 2: Solve the Quadratic Equation

We can solve the quadratic equation $x^2 + 7x + 9 = 0$ using the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

In this case, $a = 1$, $b = 7$, and $c = 9$. Plugging these values into the quadratic formula, we get:

$$x = \frac{-7 \pm \sqrt{7^2 - 4(1)(9)}}{2(1)}$$

$$x = \frac{-7 \pm \sqrt{49 - 36}}{2}$$

$$x = \frac{-7 \pm \sqrt{13}}{2}$$

So, the two solutions for x are:

$$x_1 = \frac{-7 + \sqrt{13}}{2}$$

$$x_2 = \frac{-7 - \sqrt{13}}{2}$$

These are the values of $\tan θ$ that satisfy the equation.

Step 3: Find the Angles

Now we need to find the angles θ that correspond to these values of $\tan θ$. Recall that we made the substitution $x = \tan θ$, so we have:

$$\tan θ_1 = \frac{-7 + \sqrt{13}}{2}$$

$$\tan θ_2 = \frac{-7 - \sqrt{13}}{2}$$

To find the angles, we use the inverse tangent function, $\arctan$ or $\tan^{-1}$:

$$θ_1 = \arctan\left(\frac{-7 + \sqrt{13}}{2}\right)$$

$$θ_2 = \arctan\left(\frac{-7 - \sqrt{13}}{2}\right)$$

Using a calculator, we find the approximate values:

$$θ_1 ≈ -52.9^{\circ}$$

$$θ_2 ≈ -76.4^{\circ}$$

However, these angles are not in the interval $[0^{\circ}, 360^{\circ})$. We need to find angles in this interval that have the same tangent values.

Step 4: Account for the Periodicity of Tangent

The tangent function has a period of $180^{\circ}$, which means that $\tan(θ) = \tan(θ + 180^{\circ}n)$ for any integer n. We can use this property to find angles in the interval $[0^{\circ}, 360^{\circ})$ that have the same tangent values as $θ_1$ and $θ_2$.

For $θ_1 ≈ -52.9^{\circ}$:

• Add $180^{\circ}$ to get an angle in the second quadrant: $-52.9^{\circ} + 180^{\circ} ≈ 127.1^{\circ}$
• Add another $180^{\circ}$ to get an angle in the fourth quadrant: $127.1^{\circ} + 180^{\circ} ≈ 307.1^{\circ}$

For $θ_2 ≈ -76.4^{\circ}$:

• Add $180^{\circ}$ to get an angle in the second quadrant: $-76.4^{\circ} + 180^{\circ} ≈ 103.6^{\circ}$
• Add another $180^{\circ}$ to get an angle in the fourth quadrant: $103.6^{\circ} + 180^{\circ} ≈ 283.6^{\circ}$

So, the solutions in the interval $[0^{\circ}, 360^{\circ})$ are approximately $103.6^{\circ}$, $127.1^{\circ}$, $283.6^{\circ}$, and $307.1^{\circ}$.

Step 5: Verify the Solutions

It's always a good practice to verify the solutions by plugging them back into the original equation. However, since we used approximations, the verification might not be exact, but it should be close. We can use a calculator to check the tangent values of these angles and see if they satisfy the original equation.

Conclusion

In this article, we have thoroughly explored the process of solving the trigonometric equation $\tan^2 θ + 7 \tan θ + 9 = 0$ over the interval $[0^{\circ}, 360^{\circ})$. We began by establishing a firm understanding of trigonometric equations, the tangent function, the unit circle, and the quadratic formula. We then methodically solved the equation by recognizing its quadratic form, applying the quadratic formula, finding the angles using the inverse tangent function, and accounting for the periodicity of the tangent function. The solutions we found are approximately $103.6^{\circ}$, $127.1^{\circ}$, $283.6^{\circ}$, and $307.1^{\circ}$. Solving trigonometric equations is a vital skill in mathematics and its applications. By mastering the techniques outlined in this article, you will be well-equipped to tackle a wide range of trigonometric problems. Remember, the key is to break down the problem into manageable steps, utilize the appropriate tools and concepts, and always verify your solutions.

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