Problem 31 - Precalculus Mathematics for Calculus

Question

Find all solutions of the given equation. $$3 \tan ^{2} \theta-1=0$$

Step-by-Step Solution

Answer

\(\theta = \frac{\pi}{6} + k\pi\) and \(\theta = \frac{5\pi}{6} + k\pi\), where \(k\) is any integer.

Step 1: Rearrange the Equation

Start by moving the constant term to the other side of the equation. The original equation is \(3 \tan^2 \theta - 1 = 0\). Add 1 to both sides to get \(3 \tan^2 \theta = 1\).

Step 2: Isolate \(\tan^2 \theta\)

Divide both sides by 3 to solve for \(\tan^2 \theta\). You get \(\tan^2 \theta = \frac{1}{3}\).

Step 3: Solve for \(\tan \theta\)

Take the square root of both sides of the equation. This gives \(\tan \theta = \pm \frac{1}{\sqrt{3}}\). Simplify \(\frac{1}{\sqrt{3}}\) to \(\frac{\sqrt{3}}{3}\).

Step 4: Find General Solutions Using Inverse Tangent

To find the angle \(\theta\) that corresponds to \(\tan \theta = \pm \frac{\sqrt{3}}{3}\), use the inverse tangent function: \(\theta = \tan^{-1}\left(\frac{\sqrt{3}}{3}\right) + k\pi\) and \(\theta = \tan^{-1}\left(-\frac{\sqrt{3}}{3}\right) + k\pi\), where \(k\) is any integer.

Step 5: Calculate Specific Angles

Calculate the specific solutions: \(\theta = \tan^{-1}\left(\frac{\sqrt{3}}{3}\right)\), which is \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\). For the negative case, \(\theta = \tan^{-1}\left(-\frac{\sqrt{3}}{3}\right)\) gives \(\theta = -\frac{\pi}{6}\) and \(\theta = -\frac{5\pi}{6}\).

Step 6: State General Solutions

The general solutions for \(\theta\) are given by the angles found in Step 5 with the periodic addition due to tangent's periodicity. Therefore, \(\theta = \frac{\pi}{6} + k\pi\) and \(\theta = \frac{5\pi}{6} + k\pi\) where \(k\) is any integer.

Key Concepts

Tangent Function

The tangent function, often denoted as \(\tan \theta\), is one of the main trigonometric functions and is defined as the ratio of the sine and cosine functions. Specifically, \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). This function is particularly useful in understanding angles and right triangle relationships.

• The tangent of an angle provides the slope of the line that intersects the unit circle.
• Tangent functions are periodic, meaning they repeat their values in regular intervals. The period for a tangent function is \(\pi\), or 180 degrees.

Tangent functions are commonly used in various fields like physics and engineering to model phenomena like wave patterns. When solving trigonometric equations involving tangent, the periodic nature must be taken into account, as it offers multiple solutions in different cycles.In the exercise, you first solve \(3 \tan^2 \theta = 1\), which simplifies to \(\tan \theta = \pm \frac{1}{\sqrt{3}}\). This tells us that the original equation has solutions for \(\theta\) that repeat every \(\pi\). Understanding how \(\tan \theta\) operates is crucial to solving equations correctly and in identifying all possible solutions.

Inverse Trigonometric Functions

Inverse trigonometric functions are essential when you need to determine an angle given the value of a trigonometric function. The inverse of the tangent function is denoted as \(\tan^{-1} x\), also known as arctan. This function returns the angle whose tangent is \(x\).

• The domain of \(\tan \theta\) is all real numbers, while its range for \(\tan^{-1} x\) is between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).
• To solve for \(\theta\) in the equation \(\tan \theta = x\), we use \(\theta = \tan^{-1}(x) + k\pi\).

In the given exercise, recognizing that \(\tan \theta = \pm \frac{\sqrt{3}}{3}\) points to utilizing the inverse tangent function to discover specific angles. Computing this, \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\) are found as primary solutions due to inverse tangent operations. The inclusion of \(k\pi\) accounts for all cyclic repetitions.Being able to work with inverse trigonometric functions enables reaching solutions needed for many mathematical problems where angles need to be precisely identified.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables. They are fundamental to simplifying expressions and solving equations involving trigonometric terms.

• Some basic identities include Pythagorean identity: \(\sin^2 \theta + \cos^2 \theta = 1\), and \(1 + \tan^2 \theta = \sec^2 \theta\).
• Understanding these identities can help in transforming and solving complex trigonometric equations.

In the exercise, manipulating \(3 \tan^2 \theta - 1 = 0\) utilizes the identity \(\tan^2 \theta = \frac{1}{3}\) to simplify the problem and identify the correct range of solutions. Recognizing identities and their transformations proves essential in achieving the correct answer efficiently.Trigonometric identities allow mathematicians to see beyond complicated expressions, unveiling underlying simplicity that aids in problem-solving endeavors.