Problem 3 - Algebra 2 and Trigonometry

Question

In \(3-14,\) find the exact values of \(\theta\) in the interval \(0^{\circ} \leq \theta < 360^{\circ}\) that satisfy each equation. $$ 2 \cos ^{2} \theta-3 \sin \theta=0 $$

Step-by-Step Solution

Answer

\(\theta = 30^{\circ}, 150^{\circ}\)

Step 1: Convert to Single Trigonometric Function

First, let's convert all trigonometric functions into terms of either sine or cosine.Using the Pythagorean identity, replace \ with \.\[\cos^2 \theta = 1 - \sin^2 \theta\].Substitute this into the original equation:\[2 (1 - \sin^2 \theta) - 3 \sin \theta = 0\] or simply\[2 - 2\sin^2 \theta - 3\sin \theta = 0\].

Step 2: Form Simplified Trigonometric Equation

Rearrange the equation from the previous step:\[2 - 2\sin^2 \theta - 3\sin \theta = 0\].Move all terms to one side:\[2\sin^2 \theta + 3\sin \theta - 2 = 0\].

Step 3: Factor the Quadratic Equation

Factor the quadratic form obtained in sine:\[2\sin^2 \theta + 3\sin \theta - 2 = 0 \].This factors to:\[(2\sin \theta - 1)(\sin \theta + 2) = 0 \].

Step 4: Solve Each Factor

Solve each factor for \(\sin \theta\):1. \(2\sin \theta - 1 = 0\) gives \(\sin \theta = \frac{1}{2}\).2. \(\sin \theta + 2 = 0\) gives \(\sin \theta = -2\).

Step 5: Evaluate Solutions in Boundary

Evaluate solutions in the interval \(0^{\circ} \leq \theta < 360^{\circ}\):1. The solution \(\sin \theta = \frac{1}{2}\) gives \(\theta = 30^{\circ}, 150^{\circ}\).2. Since \(\sin \theta = -2\) is not possible (as sine ranges from \(-1\) to \(1\)), it yields no solutions.

Key Concepts

Pythagorean Identity

In trigonometry, the Pythagorean Identity is a fundamental truth that relates the sine and cosine functions. It states that for any angle \( \theta \), the square of the sine of \( \theta \) plus the square of the cosine of \( \theta \) always equals 1. This is expressed mathematically as:

• \( \sin^2 \theta + \cos^2 \theta = 1 \)

This identity is crucial when solving trigonometric equations because it allows us to convert terms. For example, \( \cos^2 \theta \) can be substituted with \( 1 - \sin^2 \theta \), enabling us to rewrite equations using a single trigonometric function. This simplification was key in the original exercise to convert the equation entirely into a function of sine.

Sine Function

The sine function is one of the primary trigonometric functions, defined as the ratio of the opposite side to the hypotenuse in a right-angled triangle. For any angle \( \theta \), it's represented as \( \sin \theta \). Its value ranges from

• -1 to 1

making it especially valuable in problems involving wave patterns, oscillations, and circular motion.
The sine function is periodic, meaning it repeats its values in a regular interval known as its period, which is \(360^{\circ}\) or \(2\pi\) radians. This periodicity allows for multiple solution possibilities within a given range when solving equations like \( \sin \theta = \frac{1}{2} \), which has solutions of \( \theta = 30^{\circ}, 150^{\circ} \) within the interval \(0^{\circ} \leq \theta < 360^{\circ}\).

Cosine Function

The cosine function complements the sine function. In a right triangle, it is the ratio of the adjacent side to the hypotenuse, represented as \( \cos \theta \). Like the sine function, cosine values also range from:

• -1 to 1

which makes it equally crucial in the analysis of cycles and waveforms.
Cosine shares the same periodicity as sine, with its period also being \(360^{\circ}\) or \(2\pi\) radians. In solving equations, converting cosine functions into sine functions using the Pythagorean Identity can be strategical, as seen in changing \( \cos^2 \theta \) into \( 1 - \sin^2 \theta \) to simplify equations and adhere to a single trigonometric function. This was a vital step in bringing the original trigonometric equation to a solvable form.

Quadratic Equation

A quadratic equation is a polynomial equation of the second degree, generally expressed in the standard form as:

• \( ax^2 + bx + c = 0 \)

where \( a \), \( b \), and \( c \) are constants. In the context of trigonometric equations, these constants might correspond to coefficients involving trig functions such as \( \sin \theta \) or \( \cos \theta \).
In the exercise, after converting the original equation using the Pythagorean identity, the equation was rearranged into a quadratic form in terms of sine: \( 2\sin^2 \theta + 3\sin \theta - 2 = 0 \). Solving quadratic equations often involves factoring, where the expression is broken into simpler multipliers, or using the quadratic formula if factoring is not straightforward. This quadratic form provided a pathway to solving for \( \sin \theta \).

Factoring

Factoring is the process of expressing a polynomial as the product of its factors. It's like breaking down a number into its basic building blocks. For quadratic equations like \( ax^2 + bx + c = 0 \), factoring involves finding two binomials that multiply to the original equation.
In the exercise, the quadratic equation \( 2\sin^2 \theta + 3\sin \theta - 2 = 0 \) was successfully factored into:

• \((2\sin \theta - 1)(\sin \theta + 2) = 0\)

The fundamental step is to find values of \( \sin \theta \) that satisfy this factored form. This leads to separate simple equations: \( 2\sin \theta - 1 = 0 \) and \( \sin \theta + 2 = 0 \). Solving these leads to finding the possible values of \( \theta \) within the desired interval. Factoring streamlined the process, showing why it is a powerful tool in solving quadratic equations, including those involving trigonometric functions.