Problem 25 - Algebra 2

Question

Solve each equation for all values of \(\theta\). \(2 \sin ^{2} \theta-3 \sin \theta-2=0\)

Step-by-Step Solution

Answer

\(\theta = \frac{7\pi}{6}, \frac{11\pi}{6} \)

Step 1: Substitute Variable

Let us set a new variable for simplification: let \( x = \sin \theta \). The equation becomes \( 2x^2 - 3x - 2 = 0 \).

Step 2: Factor the Quadratic

Factor the quadratic equation \( 2x^2 - 3x - 2 \). Look for two numbers that multiply to \( 2 \times -2 = -4 \) and add to \(-3\). These numbers are \(-4\) and \(1\).

Step 3: Rewrite and Factor

Rewrite the equation as \( 2x^2 - 4x + x - 2 = 0 \). Now factor by grouping: \((2x^2 - 4x) + (x - 2) = 0\). Factor out the common terms: \(2x(x - 2) + 1(x - 2) = 0\).

Step 4: Solve for x

The expression \((2x + 1)(x - 2) = 0\) implies either \(2x + 1 = 0\) or \(x - 2 = 0\). Solving these gives \(x = -\frac{1}{2}\) or \(x = 2\). Recall \( x = \sin \theta \).

Step 5: Solve for \(\theta\) in Valid Range

The value \(x = 2\) is invalid since the sine function ranges from \(-1\) to \(1\). Solve \(\sin \theta = -\frac{1}{2}\). For \(\theta\) in the range from \(0\) to \(2\pi\), \(\theta\) can be \(\frac{7\pi}{6}\) or \(\frac{11\pi}{6}\).

Key Concepts

Quadratic Factoring

Quadratic factoring is a method used to solve polynomial equations that have a quadratic form, typically written as \( ax^2 + bx + c = 0 \). This method involves finding two numbers that multiply to give the product of the coefficient of \( x^2 \) (\( a \)) and the constant term (\( c \)), and add to give the coefficient of \( x \) (\( b \)).
For instance, in the quadratic \( 2x^2 - 3x - 2 \), we identified numbers \( -4 \) and \( 1 \) that meet these conditions:

• They multiply to \( 2 \times -2 = -4 \).
• They add up to \(-3\).

Once identified, the quadratic expression can be rewritten using these numbers to facilitate grouping and factoring. The goal is to transform the quadratic equation into a product of two binomials, which can then be set to zero to find solutions for \( x \). This factoring skill is essential as it reduces complex problems into simpler, solvable ones.

Sine Function

The sine function is a fundamental function in trigonometry, defined as the y-coordinate of a point on the unit circle as it sweeps out an angle \( \theta \) from the positive x-axis. Its range is important: it only takes values from \(-1\) to \(1\), meaning any solution must adhere to this limit. When we solve an equation like \( 2 \sin^2 \theta - 3 \sin \theta - 2 = 0 \), after substitution and factoring, it results in solutions for \( x = \sin \theta \).
In the example provided, we found \( x = -\frac{1}{2} \) and \( x = 2 \). Since \( 2 \) is outside the permissible range for sine, it is not valid. Thus, only \( x = -\frac{1}{2} \) can be accepted as a solution based on our understanding of the sine function’s range. Understanding this helps us filter and identify viable solutions in trigonometric equations.

Angle Solutions

Once the sine function outputs a value, such as \( \sin \theta = -\frac{1}{2} \), the task becomes finding the angle(s) \( \theta \) that correspond to this value. This process involves recalling knowledge about the unit circle and which angles produce specific sine values. Generally, for \( \sin \theta = -\frac{1}{2} \), the relevant angles are positioned in the third and fourth quadrants.
Within the standard range of \( 0 \) to \( 2\pi \), we can identify:

• \( \theta = \frac{7\pi}{6} \)
• \( \theta = \frac{11\pi}{6} \)

These figures correspond to where the sine function has a value of \(-\frac{1}{2}\). Recognizing patterns and knowing properties of trigonometric functions greatly aids in determining these solutions, as angles can repeat their sine values every \( 2\pi \) radians (or \( 360^\circ \) equivalent), providing multiple solutions within extended ranges.