Problem 59 - Precalculus with Limits

Question

In Exercises 59-62, use the Quadratic Formula to solve the equation in the interval \( [0, 2\pi) \). Then use a graphing utility to approximate the angle \( x \). \( 12 \sin^2 x - 13 \sin x + 3 = 0 \)

Step-by-Step Solution

Answer

Approximated solutions of the problem are \(x = \sin^{-1}(\frac{18}{24})\) and \(x = \sin^{-1}(\frac{8}{24})\).

Step 1: Identify the coefficients and apply the Quadratic Formula

Identify the coefficients in the equation: \(a = 12\), \(b = -13\), and \(c = 3\). Use these values in Quadratic Formula: \(\frac{-b \pm \sqrt{b^2-4ac}}{2a}\) to solve for \(\sin x\).

Step 2: Simplify and find values of \(\sin x\)

Plug the coefficients into the Quadratic Formula and perform the operations inside \(\sqrt{}\): \(\sin x = \frac{-(-13) \pm \sqrt{(-13)^2-4*12*3}}{2*12}\), which simplifies to: \(\sin x = \frac{13 \pm \sqrt{169-144}}{24}\), further simplifying to: \(\sin x = \frac{13 \pm 5}{24}\). So \(\sin x\) can be either \(\frac{18}{24}\) or \(\frac{8}{24}\).

Step 3: Use inverse sine function

The Quadratic Formula produced two possible solutions for \(\sin x\). Use the inverse sine function (\(\sin^{-1}\)) to find the possible values of \(x\). This gives: \(x = \sin^{-1}(\frac{18}{24})\) and \(x = \sin^{-1}(\frac{8}{24})\).

Step 4: Check the solutions

Check whether the found \(x\) solutions are in the given interval \( [0, 2\pi) \). Use a graphing utility if needed. If the solutions are outside this range, they are not valid.

Key Concepts

Trigonometric Equations

Trigonometric equations involve trigonometric functions like sine, cosine, and tangent. These equations often model periodic phenomena, such as waves. In our exercise, the equation to solve is quadratic in form:

• \(12 \sin^2 x - 13 \sin x + 3 = 0\).

To solve such an equation, we treat it like a typical quadratic equation by substituting \( \sin x\) as a variable, for example, \( y = \sin x \). This changes the equation to a standard quadratic form:

• \(12y^2 - 13y + 3 = 0\).

Using the quadratic formula lets us find the values for \(y\), which will then help us find \(x\) in terms of angles. The solution involves calculating precise values for the trigonometric function, ensuring the solutions fit into the interval needed.

Interval Notation

Interval notation is a method used to describe a range of numbers, often representing solutions or specific conditions in mathematics. In this exercise, the interval is

• \([0, 2\pi)\),

which indicates that solutions for \(x\) must be greater than or equal to 0 and less than \(2\pi\).

• The bracket \([\) means 0 is included,
• and the parenthesis \()\) means \(2\pi\) is not included.

This interval specifically represents one complete cycle of a trigonometric function. Solutions outside this range in terms of radian measure aren't valid for this problem. Make sure any solutions found fit neatly within these bounds to satisfy the condition of the interval.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as \(\sin^{-1}\), \(\cos^{-1}\), and \(\tan^{-1}\), help us find angle measures from trigonometric values. In the exercise, once we determined the possible values for \(\sin x\), we used the inverse sine function to find the angle \(x\).

• For example: \[x = \sin^{-1}\left(\frac{18}{24}\right)\] and \[x = \sin^{-1}\left(\frac{8}{24}\right)\].

Using these functions is crucial to bridge the gap between numerical results (like \(\sin x\) values) and meaningful solutions (angles in radians or degrees). Handling these functions properly ensures that the solution remains within the desired range, especially important given the interval notation.

Solving Quadratic Equations

Solving quadratic equations often involves finding "roots" or solutions that satisfy the equation. For trigonometric equations such as ours, we applied the quadratic formula:

• \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].

The coefficients \(a = 12\), \(b = -13\), \(c = 3\) were placed into the formula, resulting in the possible values for \(\sin x\).

• First, compute the discriminant: \(b^2 - 4ac = 169 - 144 = 25\).
• Then, the solutions to \(\sin x\) are determined as: \(\frac{13 \pm 5}{24}\).

This step is vital for simplifying and acquiring solutions, which then help solve the trigonometric equation. Understanding the quadratic formula's application is key to working through these problems accurately and efficiently.