Problem 34 - Precalculus with Limits

Question

In Exercises 25-38, find all solutions of the equation in the interval \( [0, 2\pi) \). \( 2 \sin^2 x + 3 \sin x + 1 = 0 \)

Step-by-Step Solution

Answer

The solutions of the equation \( 2 \sin^2 x + 3 \sin x + 1 = 0 \) in the interval [0, \(2\pi\)) are \(x = \frac{7\pi}{6}\) and \(x = \frac{11\pi}{6}\).

Step 1: Rewrite the equation

You can rewrite the equation \( 2 \sin^2 x + 3 \sin x + 1 = 0 \) as a quadratic equation in the form \( ax^2 + bx + c = 0 \). Here, let \(\sin x = y\), so the equation becomes \( 2y^2 + 3y + 1 = 0 \).

Step 2: Solve the quadratic equation

To solve the quadratic equation \( 2y^2 + 3y + 1 = 0 \), you can use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). If you plug \(a = 2\), \(b = 3\), and \(c = 1\) into the quadratic formula, you will get \( y = -1 \) and \( y = -\frac{1}{2} \).

Step 3: Back substitute

Substitute back \(\sin x = y \) to the previous step and get the solutions. The sin function has a range of \([-1, 1]\). Hence there's only one \(y\) or \(\sin x\) which falls into the domain of \(\sin x\) and that is \( \sin x = -\frac{1}{2} \).

Step 4: Find \(x\)

From the unit circle, we know that \(\sin x = -\frac{1}{2}\) at \(x = \frac{7\pi}{6}\) and \(x = \frac{11\pi}{6}\). So the solutions for the given equation in the interval [0, 2\(\pi\)) are \(x = \frac{7\pi}{6}\) and \(x = \frac{11\pi}{6}\)

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation where the highest power of the variable is squared. These equations are generally written in the standard form: \[ ax^2 + bx + c = 0 \] where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). In trigonometric contexts, sometimes you find quadratic equations in terms of trigonometric functions, like the sine function. To solve these, you typically follow these steps:

• Identify the trigonometric function as a temporary variable.
• Recast the equation to look like a typical quadratic equation.
• Solve the quadratic equation using appropriate methods like factoring, completing the square, or the quadratic formula.
• Once solved, substitute back the trigonometric function to find your solutions.

Understanding how to manipulate and solve quadratic equations is essential to tackle more complex problems like those involving trigonometric equations.

Unit Circle

The unit circle is a fundamental tool in trigonometry. It is a circle with a radius of one centered at the origin of a coordinate plane. The unit circle helps to define trigonometric functions, through the coordinates of points on the circle corresponding to angles measured in radians.The angle measured from the positive x-axis counter-clockwise to a point on the circle determines the point's coordinates:

• The x-coordinate is the cosine of the angle.
• The y-coordinate is the sine of the angle.

The unit circle allows us to determine the particular angles where specific values of sine (and cosine) occur. For example, corners of the circle, such as \( rac{7\pi}{6} \)and \( rac{11\pi}{6} \)yield \( ext{sin}(x) = -\frac{1}{2} \), helping us solve trigonometric equations.

Sine Function

The sine function is one of the primary trigonometric functions. It relates an angle in a right triangle to the ratio of the length of the opposite side to the hypotenuse. In a coordinate system, particularly the unit circle, it evaluates the y-coordinate of a point rotating around the circle.The sine function has several key properties:

• Its range is \([-1, 1]\).
• It is periodic, with a period of \(2\pi\), meaning the values repeat every \(2\pi\) radians.
• The function is odd, so \( ext{sin}(-x) = - ext{sin}(x) \).

When solving equations involving the sine function, you look for angles (or \(x\) values) that produce the desired sine values, utilizing the unit circle or inverse functions when necessary. For \( ext{sin}(x) = -\frac{1}{2} \), you find angles in specified intervals that match this condition, exploiting the periodic nature of the sine curve.