Problem 50 - Algebra and Trigonometry

Question

(a) Find all solutions of the equation. (b) Use a calculator to solve the equation in the interval \([0,2 \pi),\) correct to five decimal places. $$2 \tan x=13$$

Step-by-Step Solution

Answer

The solutions in [0, 2π) are approximately 1.3734 and 4.51499.

Step 1: Isolate Tan Function

Start with the given equation \(2 \tan x = 13\). To isolate \(\tan x\), divide both sides of the equation by 2, resulting in \(\tan x = \frac{13}{2}\).

Step 2: Find General Solutions

The general solution for \(\tan x = a\) is given by \(x = \arctan(a) + n\pi\), where \(n\) is an integer. Use this to find the solutions for \(x\). Here \(a = \frac{13}{2}\), so \(x = \arctan\left(\frac{13}{2}\right) + n\pi\).

Step 3: Use Calculator for Specific Interval

Calculate \(x = \arctan\left(\frac{13}{2}\right)\) using a calculator to get \(x \approx 1.3734\). Since the problem specifies the interval \([0, 2\pi)\), determine the possible values of \(x\) within this interval using the formula from Step 2.

Step 4: Find Solutions in Given Interval

Using \(x = 1.3734 + n\pi\), find values of \(n\) such that \(x\) remains within \([0, 2\pi)\). For \(n=0\), \(x \approx 1.3734\). For \(n=1\), \(x \approx 1.3734 + \pi \approx 4.51499\). Since \(x\) must be less than \(2\pi\), these are the solutions.

Key Concepts

Tangent Function

The tangent function is one of the six fundamental trigonometric functions. It is often abbreviated as "tan." This function takes an angle as input and returns the ratio of the opposite side to the adjacent side in a right-angled triangle. In the context of the unit circle, the tangent of an angle is equivalent to the sine of the angle divided by the cosine of the angle. Because the cosine of some angles can be zero, the tangent function has asymptotes where it is undefined.

• This typically occurs at odd multiples of \(\frac{\pi}{2}\).
• The tangent function has a periodicity of \(\pi\), meaning that it repeats its values every\(\pi\) radians.

In trigonometric equations, solving for the tangent function often requires finding angles where the tangent value repeats itself.

General Solution

A general solution in the context of trigonometric equations allows us to find all possible solutions to the equation. For equations involving the tangent function, the general solution can be expressed as \[ x = \arctan(a) + n\pi \] where \( n \) is any integer. This formula utilizes the periodic nature of the tangent function, which repeats every \(\pi\) radians.

• The term \( \arctan(a) \) gives the principal solution or the smallest positive angle \( x \) that satisfies \( \tan x = a \).
• The addition of \( n\pi \) accounts for all the "turns" or periods that can occur around the circle, hence generating all solutions.

Thus, to solve a tangent equation globally, it's essential to apply both the principal solution and the periodic nature by incorporating the generalized term.

Interval Notation

Interval notation is a concise way of representing a range of values, often used when specifying a domain or range for which a solution should be found. In trigonometric equations, it helps to narrow down the potentially infinite solutions to a specific, finite interval.The interval \([0, 2\pi)\) means all values starting from 0 up to, but not including, \(2\pi\).

• The square bracket \([\) indicates that the boundary value 0 is included.
• The parenthesis \()\) means \(2\pi\) is not included in the interval.

When solving for angles within this interval, it is essential to find solutions that stay within these boundaries. Hence, for each potential solution derived from the general formula, it must be adjusted to fit into the stated interval, ensuring all possible values are captured.

Inverse Trigonometric Functions

Inverse trigonometric functions are utilized to find angles from known trigonometric ratios. These functions essentially "reverse" what the standard trigonometric functions do. For tangent, the inverse function is denoted as \( \arctan \) or \( \tan^{-1} \).

• It accepts a ratio as input (a number) and returns an angle, typically between \(-\frac{\pi}{2}\) and \( \frac{\pi}{2}\).
• Using \( \arctan \), we can find the principal angle \( x \) such that \( \tan x \) is equal to that ratio.

Because the range of \( \arctan \) is bounded, additional solutions can be generated by considering the periodicity of the tangent function, which is \( \pi \). Therefore, when solving trigonometric equations, inverse functions are the starting point for getting the principal, or first, solution.