Problem 13 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-38: Find all solutions of the equation. $$ \sqrt{3} \tan \frac{1}{3} t=1 $$

Step-by-Step Solution

Answer

\( t = \frac{\pi}{2} + 3k\pi \) for any integer \( k \).

Step 1: Isolate Tan Function

First, we'll isolate the tangent function in the equation. We have \( \sqrt{3} \tan\left( \frac{1}{3} t \right) = 1 \). Divide both sides by \( \sqrt{3} \) to isolate the tangent function: \( \tan\left( \frac{1}{3} t \right) = \frac{1}{\sqrt{3}} \).

Step 2: Identify Reference Angle

We know that \( \tan\left( \theta \right) = \frac{1}{\sqrt{3}} \) corresponds to \( \theta = \frac{\pi}{6} \) in the first quadrant.

Step 3: General Solution for Tangent

The general solution for \( \tan(\theta) = \tan(\frac{\pi}{6}) \) is \( \theta = \frac{\pi}{6} + k\pi \), where \( k \) is any integer.

Step 4: Replace \( \theta \) with \( \frac{1}{3} t \)

Substitute back to find \( t \). Replace \( \theta \) with \( \frac{1}{3} t \), giving \( \frac{1}{3} t = \frac{\pi}{6} + k\pi \).

Step 5: Solve for \( t \)

Multiply both sides by 3 to solve for \( t \): \( t = \frac{\pi}{2} + 3k\pi \).

Step 6: Conclusion

The solution to the equation is \( t = \frac{\pi}{2} + 3k\pi \) for any integer \( k \).

Key Concepts

Tangent Function

The tangent function, denoted as \( \tan(\theta) \), is one of the fundamental functions in trigonometry. It relates an angle in a right triangle to the ratio of the opposite side over the adjacent side. In terms of coordinates, for any angle \( \theta \) on the unit circle, the value of the tangent function is the quotient of the sine and cosine functions: \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \).

This function has certain key characteristics:

• Periodicity: The tangent function is periodic with a period of \( \pi \), which means its values repeat every \( \pi \) radians.
• Asymptotes: It has vertical asymptotes where the cosine function equals zero, as the tangent function becomes undefined at these points.
• Range: The range of the tangent function is all real numbers, unlike sine and cosine which are bounded between -1 and 1.

Understanding the properties of the tangent function is crucial when solving trigonometric equations. It plays a pivotal role in determining solutions within specific intervals.

Reference Angle

A reference angle is a tool used in trigonometry to simplify the process of finding the angle whose trigonometric function equals a given value. For any given angle in standard position, the reference angle is the smallest angle between the terminal side of the given angle and the x-axis.

Reference angles help in finding the trigonometric values for angles that fall into different quadrants. This is because:

• The trigonometric values of an angle correspondingly reflect across axes within the same quadrant settings.
• In the first quadrant, the angle itself is the reference angle.

In the exercise, we identified the reference angle \( \theta \) by recognizing that \( \tan(\theta) = \frac{1}{\sqrt{3}} \) corresponds to the well-known value \( \theta = \frac{\pi}{6} \). This reference angle enables determining equivalent angles in all its periodicity and other quadrants.

General Solution

The general solution for trigonometric equations refers to expressing all possible solutions in terms of some parameter, usually an integer. For the tangent function, due to its periodic nature, solutions repeat every \( \pi \) radians.

The general solution of an equation like \( \tan(\theta) = \tan(A) \) can be written as \( \theta = A + k\pi \), where \( A \) is a particular solution and \( k \) is any integer. This accounts for the infinite number of solutions across the full range of angles.

Finding the general solution is a key step in solving trigonometric equations, as it allows you to express all the possible angle values that satisfy the equation. In this task, it is crucial to understand that the general solution reflects all such angles, not just those in a limited range or specific interval.

Solving Equations

Solving trigonometric equations involves several key steps. It typically starts with isolating the trigonometric function before proceeding to find the angle(s) that satisfy the equation.Here's the general approach used to solve the given exercise:

• Isolate the Function: Before you can solve for an angle, the equation must be simplified in a way that the trigonometric function sits alone on one side. We began with \( \sqrt{3} \tan\left(\frac{1}{3} t \right) = 1 \) and isolated \( \tan\left(\frac{1}{3}t\right) \) by dividing by \( \sqrt{3} \).
• Determine the Reference Angle: Identify known angles where the trigonometric function equals the right-hand side of the equation. In our equation, this was \( \theta = \frac{\pi}{6} \).
• Apply the General Solution: Due to the repetitive nature of trigonometric functions, express the solutions in parameterized form. For tangent equations, this is \( \theta + k \pi \).
• Solve for the Variable: Once you have the general form, solve for the required variable by substituting back. Here, we substituted \( \theta = \frac{1}{3} t \) back and solved for \( t \).

By following these steps methodically, you ensure that no potential solutions are missed, thus thoroughly solving the equation for all possible values.