Problem 6 - Algebra 2 and Trigonometry

Question

In \(3-14\) , use the quadratic formula to find, to the nearest degree, all values of \(\theta\) in the interval \(0^{\circ} \leq \theta<360^{\circ}\) that satisfy each equation. $$ 9 \sin ^{2} \theta+6 \sin \theta=2 $$

Step-by-Step Solution

Answer

The values of \(\theta\) are \(14^{\circ}\) and \(166^{\circ}\).

Step 1: Identify the equation in standard quadratic form

The given equation is \(9 \sin^2 \theta + 6 \sin \theta = 2\). This can be rearranged to the standard quadratic form as \(9 \sin^2 \theta + 6 \sin \theta - 2 = 0\). The equation is in terms of \(\sin \theta\), where \(a = 9\), \(b = 6\), and \(c = -2\).

Step 2: Use the quadratic formula

The quadratic formula is \( \sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Plugging in the values of \(a\), \(b\), and \(c\), we get \( \sin \theta = \frac{-6 \pm \sqrt{6^2 - 4 \times 9 \times (-2)}}{2 \times 9} \).

Step 3: Calculate the discriminant and roots

Calculate the discriminant: \(b^2 - 4ac = 6^2 - 4 \times 9 \times (-2) = 36 + 72 = 108\). Now compute the square root of the discriminant: \(\sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3}\). Substitute back into the quadratic formula to find: \( \sin \theta = \frac{-6 \pm 6\sqrt{3}}{18}\). Simplify the roots to get: \( \sin \theta = \frac{-1 \pm \sqrt{3}}{3}\).

Step 4: Determine the valid sine values

The possible solutions for \(\sin \theta\) are \(\sin \theta = \frac{-1 + \sqrt{3}}{3}\) and \(\sin \theta = \frac{-1 - \sqrt{3}}{3}\). Evaluate these numerically: \(\frac{-1 + \sqrt{3}}{3} \approx 0.244\) and \(\frac{-1 - \sqrt{3}}{3} < -1\), which is invalid as \(\sin \theta\) must be between \(-1\) and \(1\).

Step 5: Find angles for valid \(\sin \theta\)

Consider \(\sin \theta = 0.244\) which is valid. The solution exists in the first and second quadrants. Calculate: \(\theta_1 \approx \arcsin(0.244) \approx 14.1^{\circ}\) and \(\theta_2 = 180^{\circ} - 14.1^{\circ} = 165.9^{\circ}\), rounding to the nearest degree gives \(\theta = 14^{\circ}\) and \(\theta = 166^{\circ}\).

Key Concepts

Understanding the Sine Function

The sine function is fundamental in trigonometry, representing the y-coordinate of a point on the unit circle as the angle in radians or degrees changes. It describes how the height of a point above the horizontal axis varies as we move around the circle.
For any angle \( \theta \), the sine function gives us \( \sin(\theta) \), which ranges between -1 and 1. This is crucial because any acceptable solution for \( \sin \theta \) must lie within these bounds.
In the unit circle:

• \( \sin(0^\circ) = 0 \)
• \( \sin(90^\circ) = 1 \)
• \( \sin(180^\circ) = 0 \)
• \( \sin(270^\circ) = -1 \)

Understanding this range and behavior helps in solving trigonometric equations where we equate a function of \( \sin \theta \) to some value. Visualizing this as a wave—cyclical and periodic—can aid in grasping how angles relate to specific sine values.

Solving a Trigonometric Equation

A trigonometric equation often involves trigonometric functions like sine, cosine, or tangent set equal to a value. The goal is to find all the angle measures within a specified interval that satisfy this equation. For example, in our exercise, we have the equation \[9 \sin^2 \theta + 6 \sin \theta - 2 = 0,\]which breaks down into working with \( \sin \theta \).
This equation mirrors the structure of a quadratic equation \( ax^2 + bx + c = 0 \). Identifying this helps us apply algebraic techniques, such as the quadratic formula, traditionally used on quadratic expressions.
To solve:

• Express it in standard form: \(9 \sin^2 \theta + 6 \sin \theta - 2 = 0\).
• Isolate \( \sin \theta \) using the quadratic formula.
• Evaluate resulting values to see which are within the acceptable range of sine.

Once the values of \( \sin \theta \) are found within \([-1, 1]\), the corresponding angles are determined. Typically, the calculator's \( \arcsin \) or inverse sine function provides principal angles, while additional solutions often arise due to the sine function's periodic nature.

Discriminant Calculation in Quadratic Equations

The discriminant of a quadratic equation \( ax^2 + bx + c = 0 \) is given by \( b^2 - 4ac \). This value is crucial because it determines the nature and number of roots or solutions of the equation.
Here's how the discriminant impacts solutions:

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is exactly one real root (repeated).
• If negative, there are no real roots—only complex ones.

For the given equation \(9 \sin^2 \theta + 6 \sin \theta - 2 = 0\), we calculate:\[b^2 - 4ac = 6^2 - 4 \times 9 \times (-2) = 36 + 72 = 108.\]This positive discriminant indicates two distinct real roots. These roots correspond to possible values of \( \sin \theta \), which we then evaluate to confirm if they lie within the required range of sine values. If they do, they lead us to the valid angles or solutions for the original trigonometric equation.