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Ch. 3 - Trigonometric Identities and Equations
Blitzer - Trigonometry 3rd Edition
Blitzer3rd EditionTrigonometryISBN: 9780137316601Not the one you use?Change textbook
Chapter 3, Problem 89

In Exercises 85–96, use a calculator to solve each equation, correct to four decimal places, on the interval [0, 2𝝅). tan x = ﹣3

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1
Recognize that the equation to solve is \(\tan x = -3\) on the interval \([0, 2\pi)\), where \(x\) is the angle in radians.
Recall that the tangent function has a period of \(\pi\), so solutions will repeat every \(\pi\) radians.
Use the inverse tangent function to find the principal value: calculate \(x_1 = \arctan(-3)\) using a calculator. This will give an angle in the range \((-\frac{\pi}{2}, \frac{\pi}{2})\).
Since \(\tan x\) is negative, identify the quadrants where tangent is negative. Tangent is negative in the second and fourth quadrants, so find the corresponding angles in \([0, 2\pi)\) by adding \(\pi\) to \(x_1\) or using symmetry.
Write the two solutions in the interval \([0, 2\pi)\) as \(x_1\) (adjusted to be positive if necessary) and \(x_2 = x_1 + \pi\). These two values are the solutions to \(\tan x = -3\) on the given interval.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Understanding the Tangent Function

The tangent function, tan(x), is the ratio of the sine and cosine functions (sin x / cos x). It is periodic with period π, meaning tan(x + π) = tan x. Knowing its behavior and range helps in solving equations involving tan x within a specified interval.
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Using a Calculator to Find Inverse Tangent

To solve tan x = -3, use the inverse tangent function (arctan or tan⁻¹) on a calculator to find the principal value. Since the tangent function is periodic and can be negative in certain quadrants, additional solutions must be found by considering the function's period and sign.
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Solving Trigonometric Equations on a Given Interval

When solving tan x = -3 on [0, 2π), identify all solutions within one full cycle. Because tan x has period π, there will be two solutions in [0, 2π). Adjust the principal value by adding π to find the second solution, and round answers to four decimal places as required.
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