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7.4 Basic Trigonometric Equations Copyright © Cengage Learning. All rights reserved.

Objectives Basic Trigonometric Equations Solving Trigonometric Equations by Factoring

Basic Trigonometric Equations

Basic Trigonometric Equations An equation that contains trigonometric functions is called a trigonometric equation. For example, the following are trigonometric equations: sin2  + cos2  = 1 2 sin – 1 = 0 tan 2 – 1 = 0 The first equation is an identity—that is, it is true for every value of the variable . The other two equations are true only for certain values of . To solve a trigonometric equation, we find all the values of the variable that make the equation true.

Basic Trigonometric Equations Solving any trigonometric equation always reduces to solving a basic trigonometric equation—an equation of the form T( ) = c, where T is a trigonometric function and c is a constant. In the next examples we solve such basic equations.

Example 1 – Solving a Basic Trigonometric Equation Solve the equation Solution: Find the solutions in one period. Because sine has period 2, we first find the solutions in any interval of length 2. To find these solutions, we look at the unit circle in Figure 1. Figure 1

Example 1 – Solution cont’d We see that sin  = in Quadrants I and II, so the solutions in the interval [0, 2) are Find all solutions. Because the sine function repeats its values every 2 units, we get all solutions of the equation by adding integer multiples of 2 to these solutions: where k is any integer.

Example 1 – Solution cont’d Figure 2 gives a graphical representation of the solutions. Figure 2

Example 4 – Solving a Basic Trigonometric Equation Solve the equation tan  = 2. Solution: Find the solutions in one period. We first find one solution by taking tan–1 of each side of the equation. tan  = 2  = tan–1(2)   1.12 Given equation Take tan–1 of each side Calculator (in radian mode)

Example 4 – Solution cont’d By the definition of tan–1 the solution that we obtained is the only solution in the interval (– /2,  /2) (which is an interval of length ). Find all solutions. Since tangent has period , we get all solutions of the equation by adding integer multiples of  :   1.12 + k where k is any integer.

Example 4 – Solution cont’d A graphical representation of the solutions is shown in Figure 6. You can check that the solutions shown in the graph correspond to k = –1, 0, 1, 2, 3. Figure 6

Basic Trigonometric Equations In the next example we solve trigonometric equations that are algebraically equivalent to basic trigonometric equations.

Example 5 – Solving Trigonometric Equations Find all solutions of the equation. (a) 2 sin  – 1 = 0 (b) tan2  – 3 = 0 Solution: (a) We start by isolating sin  . 2 sin  – 1 = 0 2 sin  = 1 sin  = Given equation Add 1 Divide by 2

Example 5 – Solution cont’d This last equation is the same as that in Example 1. The solutions are  = + 2k  = + 2k where k is any integer. (b) We start by isolating tan . tan2  – 3 = 0 tan2  = 3 tan  = Given equation Add 3 Take the square root

Example 5 – Solution cont’d Because tangent has period , we first find the solutions in any interval of length . In the interval (– /2,  /2) the solutions are  =  /3 and  = – /3. To get all solutions, we add integer multiples of  to these solutions:  = + k  = – + k where k is any integer.

Solving Trigonometric Equations by Factoring

Solving Trigonometric Equations by Factoring Factoring is one of the most useful techniques for solving equations, including trigonometric equations. The idea is to move all terms to one side of the equation, factor, and then use the Zero-Product Property.

Example 6 – A Trigonometric Equation of Quadratic Type Solve the equation 2 cos2  – 7 cos  + 3 = 0. Solution: We factor the left-hand side of the equation. 2 cos2  – 7 cos  + 3 = 0 (2 cos  – 1)(cos  – 3) = 0 2 cos  – 1 = 0 or cos  – 3 = 0 cos  = or cos  = 3 Given equation Factor Set each factor equal to 0 Solve for cos 

Example 6 – Solution cont’d Because cosine has period 2, we first find the solutions in the interval [0, 2). For the first equation the solutions are  =  /3 and  = 5 /3 (see Figure 7). Figure 7

Example 6 – Solution cont’d The second equation has no solution because cos  is never greater than 1. Thus the solutions are  = + 2k  = + 2k where k is any integer.

Example 7 – Solving a Trigonometric Equation by Factoring Solve the equation 5 sin  cos  + 4 cos  = 0. Solution: We factor the left-hand side of the equation. 5 sin  cos  + 2 cos  = 0 cos  (5 sin  + 2) = 0 cos  = 0 or 5 sin  + 4 = 0 sin  = –0.8 Given equation Factor Set each factor equal to 0 Solve for sin 

Example 7 – Solution cont’d Because sine and cosine have period 2, we first find the solutions of these equations in an interval of length 2. For the first equation the solutions in the interval [0, 2) are  =  /2 and  = 3 /2 . To solve the second equation, we take sin–1 of each side. sin  = –0.80  = sin–1(–0.80) Second equation Take sin–1 of each side

Example 7 – Solution cont’d   –0.93 So the solutions in an interval of length 2 are  = –0.93 and  =  + 0.93  4.07 (see Figure 8). Calculator (in radian mode) Figure 8

Example 7 – Solution cont’d We get all the solutions of the equation by adding integer multiples of 2 to these solutions.  = + 2k  = + 2k   –0.93 + 2k   4.07 + 2k where k is any integer.