Analytic Trigonometry

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Analytic Trigonometry 5 Analytic Trigonometry Copyright © Cengage Learning. All rights reserved.

5.3 Solving Trigonometric Equations Copyright © Cengage Learning. All rights reserved.

Objectives Use standard algebraic techniques to solve trigonometric equations. Solve trigonometric equations of quadratic type. Solve trigonometric equations involving multiple angles. Use inverse trigonometric functions to solve trigonometric equations.

Introduction

Introduction To solve a trigonometric equation, use standard algebraic techniques (when possible) such as collecting like terms and factoring. Your preliminary goal in solving a trigonometric equation is to isolate the trigonometric function on one side of the equation. For example, to solve the equation 2 sin x = 1, divide each side by 2 to obtain

Introduction To solve for x, note in the figure below that the equation has solutions x =  /6 and x = 5 /6 in the interval [0, 2).

Introduction Moreover, because sin x has a period of 2, there are infinitely many other solutions, which can be written as and where n is an integer, as shown above. General solution

Introduction The figure below illustrates another way to show that the equation has infinitely many solutions. Any angles that are coterminal with  /6 or 5 /6 will also be solutions of the equation. When solving trigonometric equations, you should write your answer(s) using exact values rather than decimal approximations.

Example 1 – Collecting Like Terms Solve Solution: Begin by isolating sin x on one side of the equation. Write original equation. Add sin x to each side. Subtract from each side.

Example 1 – Solution cont’d Combine like terms. Because sin x has a period of 2, first find all solutions in the interval [0, 2). These solutions are x = 5 /4 and x = 7 /4. Finally, add multiples of 2 to each of these solutions to obtain the general form and where n is an integer. Divide each side by 2. General solution

Equations of Quadratic Type

Equations of Quadratic Type Many trigonometric equations are of quadratic type ax2 + bx + c = 0, as shown below. Quadratic in sin x Quadratic in sec x 2 sin2 x – sin x – 1 = 0 sec2 x – 3 sec x – 2 = 0 2(sin x)2 – sin x – 1 = 0 (sec x)2 – 3(sec x) – 2 = 0 To solve equations of this type, factor the quadratic or, when this is not possible, use the Quadratic Formula.

Example 4 – Factoring an Equation of Quadratic Type Find all solutions of 2 sin2 x – sin x – 1 = 0 in the interval [0, 2). Solution: Treat the equation as a quadratic in sin x and factor. 2 sin2 x – sin x – 1 = 0 (2 sin x + 1)(sin x – 1) = 0 Write original equation. Factor.

Example 4 – Solution cont’d Setting each factor equal to zero, you obtain the following solutions in the interval [0, 2). 2 sin x + 1 = 0 and sin x – 1 = 0

Functions Involving Multiple Angles

Functions Involving Multiple Angles The next example involve trigonometric functions of multiple angles of the forms cos ku and tan ku. To solve equations of these forms, first solve the equation for ku, and then divide your result by k.

Example 7 – Solving a Multiple-Angle Equation Solve 2 cos 3t – 1 = 0. Solution: 2 cos 3t – 1 = 0 2 cos 3t = 1 Write original equation. Add 1 to each side. Divide each side by 2.

Example 7 – Solution cont’d In the interval [0, 2), you know that 3t =  /3 and 3t = 5 /3 are the only solutions, so, in general, you have and Dividing these results by 3, you obtain the general solution where n is an integer. General solution

Using Inverse Functions

Example 9 – Using Inverse Functions sec2 x – 2 tan x = 4 1 + tan2 x – 2 tan x – 4 = 0 tan2 x – 2 tan x – 3 = 0 (tan x – 3)(tan x + 1) = 0 Setting each factor equal to zero, you obtain two solutions in the interval (– /2,  /2). [Recall that the range of the inverse tangent function is (– /2,  /2).] x = arctan 3 and x = arctan(–1) = – /4 Original equation. Pythagorean identity Combine like terms. Factor.

Example 9 – Using Inverse Functions cont’d Finally, because tan x has a period of , you add multiples of  to obtain x = arctan 3 + n and x = (– /4 ) + n where n is an integer. You can use a calculator to approximate the value of arctan 3. General solution