Problem 33 - Algebra 2

Question

Solve each equation for \(0 \leq \theta<2 \pi\). $$ 2 \sin ^{2} \theta-3 \sin \theta=2 $$

Step-by-Step Solution

Answer

The solution for \(2 \sin ^{2} \theta - 3 \sin \theta = 2\) in range \(0 \leq \theta<2 \pi\) is \(\theta = 7\pi/6, 11\pi/6\).

Step 1: Rewrite as quadratic equation

The given equation \(2 \sin ^{2} \theta - 3 \sin \theta - 2 = 0\) can be regarded as quadratic equation \(2x^2 - 3x - 2 = 0\) where \(x = sin\theta\).

Step 2: Solve for x using quadratic formula

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). For the equation \(2x^2 - 3x - 2 = 0\), a = 2, b = -3 and c = -2. Substituting these values into the formula gives \(x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4*2*(-2)}}{2*2}\), which simplifies to \(x = \frac{3 \pm \sqrt{9 + 16}}{4}\) to \(x = \frac{3 \pm 5}{4}\), yielding two solutions \(x = 2\) and \(x = -0.5\).

Step 3: Calculate \( \theta \)

Since \(x = sin\theta\), the solutions \(x = 2\) and \(x = -0.5\) provide two equations \(\sin\theta = 2\) and \(\sin\theta = -0.5\). However, sine function values lie between [-1, 1], so \(\sin\theta = 2\) has no solution. For the \( \sin\theta = -0.5 \) equation, use the inverse sine function to find \(\theta = \arcsin(-0.5)\). Remember, the range for \(\theta\) is \(0 \leq \theta<2 \pi\), so the solutions are \(\theta = 7\pi/6\) and \( \theta = 11\pi/6\). These are the values of \(\theta\) in the given range that make \(\sin\theta = -0.5\).

Key Concepts

Quadratic equations

Quadratic equations are fundamental in mathematics, especially when dealing with polynomial expressions. A quadratic equation is typically expressed in the form: \[ ax^2 + bx + c = 0 \]where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable. In this context, the equation involves the sine function, where \( x = \sin\theta \), transforming the equation into \[ 2\sin^2\theta - 3\sin\theta - 2 = 0 \]Understanding how to reformulate equations as quadratics allows us to use well-established methods to find solutions.

• This equation is solved using the quadratic formula, a general solution method for any quadratic equation.
• The formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]provides the possible values for \( x \), taking into account the coefficients \( a \), \( b \), and \( c \).
• Using these values helps in determining sine function solutions.

Sine function

The sine function is one of the primary trigonometric functions, defined using the ratios in a right-angled triangle or on the unit circle. For a given angle \( \theta \), the sine function is represented as \( \sin\theta \) and typically ranges between -1 and 1. This inherent property is crucial when solving equations involving sine.

• When \( \sin\theta = 2 \), this presents an impossible scenario since the range of the sine function doesn't exceed 1.
• In contrast, \( \sin\theta = -0.5 \) is valid and requires solving for \( \theta \).
• Knowledge of the sine function's periodic nature is also key, as it helps determine multiple solutions within a specified interval.

Unit circle

The unit circle is an essential concept in trigonometry, providing a geometric interpretation of trigonometric functions. It is a circle with a radius of 1, centered at the origin of a coordinate plane. The unit circle facilitates finding values of trigonometric functions for various angles.

• From the unit circle, \( \sin\theta \) represents the y-coordinate of an angle \( \theta \) drawn from the positive x-axis.
• To solve \( \sin\theta = -0.5 \), identify the angles where the y-coordinate of the unit circle equals -0.5.
• Within the interval \( 0 \leq \theta < 2\pi \), these angles are \( 7\pi/6 \) and \( 11\pi/6 \), providing two valid solutions.

Inverse trigonometric functions

Inverse trigonometric functions help us find angles when given trigonometric ratios. Specifically, the inverse of the sine function, written as \( \arcsin \), retrieves an angle whose sine is a given number.

• Since \( \sin\theta = -0.5 \), the function \( \arcsin(-0.5) \) begins the process of determining the potential values of \( \theta \).
• Typically, the range of \( \arcsin(x) \) is from \(-\pi/2\) to \(\pi/2\). When \( x = -0.5 \), the principal value is \(-\pi/6\).
• However, considering the required range of \( \theta \) from \( 0 \) to \( 2\pi \), the solutions appropriately shift to \( 7\pi/6 \) and \( 11\pi/6 \), falling within the cycle of the sine wave on the unit circle.