Problem 24 - Precalculus with Calculus Previews

Question

Find all solutions of the given trigonometric equation if \(x\) is a real number and \(\theta\) is an angle measured in degrees. $$ 2 \sin ^{2} \theta-\sin \theta-1=0 $$

Step-by-Step Solution

Answer

Solutions: \(\theta = 90^\circ + 360^\circ k\), \(\theta = 210^\circ + 360^\circ k\), and \(\theta = 330^\circ + 360^\circ k\), where \(k\) is an integer.

Step 1: Recognize the Equation Type

The given equation \(2 \sin^2 \theta - \sin \theta - 1 = 0\) is a quadratic equation in terms of \(\sin \theta\). It can be rewritten as \(a \sin^2 \theta + b \sin \theta + c = 0\) where \(a = 2\), \(b = -1\), and \(c = -1\).

Step 2: Apply the Quadratic Formula

To solve for \(\sin \theta\), we can use the quadratic formula: \(\sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substituting \(a = 2\), \(b = -1\), and \(c = -1\), we get:\[\sin \theta = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 8}}{4}= \frac{1 \pm 3}{4}\]

Step 3: Simplify the Results

Calculate the discriminant \(\sqrt{9} = 3\). This gives us two potential solutions for \(\sin \theta\):1. \(\sin \theta = \frac{1 + 3}{4} = 1\) 2. \(\sin \theta = \frac{1 - 3}{4} = -\frac{1}{2}\)

Step 4: Determine Solutions in Degrees

For \(\sin \theta = 1\), the angle \(\theta = 90^\circ\).For \(\sin \theta = -\frac{1}{2}\), the possible angles in the interval \([0^\circ, 360^\circ)\) are \(210^\circ\) and \(330^\circ\), since sine is negative in the third and fourth quadrants.

Step 5: Express General Solutions

For all real number solutions, express \(\theta\) as:1. \(\theta = 90^\circ + 360^\circ k\) for \(\sin \theta = 1\).2. \(\theta = 210^\circ + 360^\circ k\) and \(\theta = 330^\circ + 360^\circ k\) for \(\sin \theta = -\frac{1}{2}\), where \(k\) is any integer.

Key Concepts

Quadratic Formula

The quadratic formula is a crucial tool for solving equations of the form \( ax^2 + bx + c = 0 \). In this exercise, our equation is quadratic in nature, expressed as \( 2 \sin^2 \theta - \sin \theta - 1 = 0 \). We equate \( \sin \theta \) to \( x \), transforming the equation into \( 2x^2 - x - 1 = 0 \). The quadratic formula is given as: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]By substituting \( a = 2 \), \( b = -1 \), and \( c = -1 \), it becomes:

• \( x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \)
• \( x = \frac{1 \pm 3}{4} \)

This results in potential solutions of \( x = 1 \) and \( x = -\frac{1}{2} \). By replacing \( x \) with \( \sin \theta \), we find the corresponding solutions are \( \sin \theta = 1 \) and \( \sin \theta = -\frac{1}{2} \).

Sine Function

The sine function is one of the primary trigonometric functions and plays a key role in this problem. Defined for an angle \( \theta \), \( \sin \theta \) represents the ratio of the length of the opposite side to the hypotenuse in a right-angled triangle. The range of the sine function is between -1 and 1.
It's periodic with a period of 360 degrees, which means it repeats values every full circle around the unit circle. In the context of this exercise, the sine function helps determine the angles for which the trigonometric equation is satisfied.

• For \( \sin \theta = 1 \), \( \theta = 90^\circ \).
• For \( \sin \theta = -\frac{1}{2} \), \( \theta \) values are found in the third and fourth quadrants of the unit circle: \( \theta = 210^\circ \) and \( \theta = 330^\circ \).

Angle Measurement

Angles can be measured in degrees or radians. In this problem, we're utilizing degrees, which divides a full circle into 360 units. Understanding angle measurements is critical for solving trigonometric problems because they determine the position on the unit circle.Trigonometric equations often involve finding angles that satisfy certain sine, cosine, or tangent values. The given equation arises from substituting values back into sine to find related angle measures.

• \( \theta = 90^\circ \) for \( \sin \theta = 1 \).
• \( \theta = 210^\circ \) and \( \theta = 330^\circ \) for \( \sin \theta = -\frac{1}{2} \).

Knowing how to read and convert angles in degrees helps identify angles where a given trigonometric moment occurs.

General Solutions

To fully express solutions for trigonometric equations, we often consider the general form that incorporates all possible angles a particular sine or cosine value can yield. The general solution accounts for the periodic nature of trigonometric functions:

• For \( \sin \theta = 1 \), the general solution is \( \theta = 90^\circ + 360^\circ k \).
• For \( \sin \theta = -\frac{1}{2} \), general solutions are \( \theta = 210^\circ + 360^\circ k \) and \( \theta = 330^\circ + 360^\circ k \).

Here, \( k \) represents any integer and captures the continuous nature of the function as it repeats every 360 degrees. By considering \( k \), we ensure all solutions are covered for any periodic interval.