Solving Trigonometric Equations

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1 OpenStax-CNX module: m Solving Trigonometric Equations OpenStax College This work is produced by OpenStax-CNX and licensed under the Creative Commons Attribution License 4.0 In this section, you will: Abstract Solve linear trigonometric equations in sine and cosine. Solve equations involving a single trigonometric function. Solve trigonometric equations using a calculator. Solve trigonometric equations that are quadratic in form. Solve trigonometric equations using fundamental identities. Solve trigonometric equations with multiple angles. Solve right triangle problems. Figure 1: Egyptian pyramids standing near a modern city. (credit: Oisin Mulvihill) Thales of Miletus (circa BC) is known as the founder of geometry. The legend is that he calculated the height of the Great Pyramid of Giza in Egypt using the theory of similar triangles, which he developed by measuring the shadow of his sta. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the angle of elevation and the angle of depression are found using similar triangles. Version 1.6: Jan 16, 015 1:5 pm

2 OpenStax-CNX module: m49398 In earlier sections of this chapter, we looked at trigonometric identities. Identities are true for all values in the domain of the variable. In this section, we begin our study of trigonometric equations to study real-world scenarios such as the nding the dimensions of the pyramids. 1 Solving Linear Trigonometric Equations in Sine and Cosine Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specic values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specied interval. However, just as often, we will be asked to nd all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an innite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The period of both the sine function and the cosine function is. In other words, every units, the y-values repeat. If we need to nd all possible solutions, then we must add k, where k is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is : sin θ = sin (θ ± k) (1) There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, nd common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections. Example 1 Solving a Linear Trigonometric Equation Involving the Cosine Function Find all possible exact solutions for the equation cos θ = 1. From the unit circle, we know that cos θ = 1 θ = 3, 5 3 These are the solutions in the interval [0, ]. All possible solutions are given by () θ = 3 ± k and θ = 5 3 ± k (3) where k is an integer. Example Solving a Linear Equation Involving the Sine Function Find all possible exact solutions for the equation sin t = 1. Solving for all possible values of t means that solutions include angles beyond the period of. From, we can see that the solutions are t = 6 and t = 5 6. But the problem is asking for all possible values that solve the equation. Therefore, the answer is

3 OpenStax-CNX module: m t = 6 ± k and t = 5 6 ± k (4) where k is an integer. How To: Given a trigonometric equation, solve using algebra. 1.Look for a pattern that suggests an algebraic property, such as the dierence of squares or a factoring opportunity..substitute the trigonometric expression with a single variable, such as x or u. 3.Solve the equation the same way an algebraic equation would be solved. 4.Substitute the trigonometric expression back in for the variable in the resulting expressions. 5.Solve for the angle. Example 3 Solve the Trigonometric Equation in Linear Form Solve the equation exactly: cos θ 3 = 5, 0 θ <. Use algebraic techniques to solve the equation. cos θ 3 = 5 cos θ = cos θ = 1 θ = (5) Try It: Exercise 4 ( on p. 4.) Solve exactly the following linear equation on the interval [0, ) : sin x + 1 = 0. Solving Equations Involving a Single Trigonometric Function When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see ). We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly dierent from one containing a sine or cosine function. First, as we know, the period of tangent is, not. Further, the domain of tangent is all real numbers with the exception of odd integer multiples of,unless, of course, a problem places its own restrictions on the domain.

4 OpenStax-CNX module: m Example 4 Solving a Problem Involving a Single Trigonometric Function Solve the problem exactly: sin θ 1 = 0, 0 θ <. As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate sinθ. Then we will nd the angles. sin θ 1 = 0 sin θ = 1 sin θ = 1 sin θ = ± 1 sin θ = ± 1 = ± θ = 4, 3 4, 5 4, 7 4 (6) Example 5 Solving a Trigonometric Equation Involving Cosecant Solve the following equation exactly: csc θ =, 0 θ < 4. We want all values of θ for which csc θ = over the interval 0 θ < 4. csc θ = 1 sin θ = sin θ = 1 θ = 7 6, 11 6, 19 6, 3 6 (7) Analysis As sin θ = 1, notice that all four solutions are in the third and fourth quadrants. Example 6 Solving an Equation Involving Tangent Solve the equation exactly: tan ( θ ) = 1, 0 θ <. Recall that the tangent function has a period of. On the interval [0, ), and at the angle of 4, the tangent has a value of 1. However, the angle we want is ( θ ). Thus, if tan ( 4 ) = 1,then Over the interval [0, ), we have two solutions: θ = 4 θ = 3 4 ± k (8)

5 OpenStax-CNX module: m θ = 3 4 and θ = = 7 4 (9) Try It: Exercise 8 ( on p. 4.) Find all solutions for tan x = 3. Example 7 Identify all s to the Equation Involving Tangent Identify all exact solutions to the equation (tan x + 3) = 5 + tan x, 0 x <. We can solve this equation using only algebra. Isolate the expression tan x on the left side of the equals sign. (tanx) + (3) = 5 + tanx tan x + 6 = 5 + tan x tanx tanx = 5 6 tanx = 1 (10) There are two angles on the unit circle that have a tangent value of 1 : θ = 3 4 and θ = Solve Trigonometric Equations Using a Calculator Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem. Example 8 Using a Calculator to Solve a Trigonometric Equation Involving Sine Use a calculator to solve the equation sin θ = 0.8, where θ is in radians. Make sure mode is set to radians. To nd θ, use the inverse sine function. On most calculators, you will need to push the ND button and then the SIN button to bring up the sin 1 function. What is shown on the screen issin 1 (.The calculator is ready for the input within the parentheses. For this problem, we enter sin 1 (0.8), and press ENTER. Thus, to four decimals places, The solution is sin 1 (0.8) (11)

6 OpenStax-CNX module: m The angle measurement in degrees is θ ± k (1) θ 53.1 θ (13) Analysis Note that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine. The other angle is obtained by using θ. Example 9 Using a Calculator to Solve a Trigonometric Equation Involving Secant Use a calculator to solve the equation sec θ = 4, giving your answer in radians. We can begin with some algebra. sec θ = 4 1 cos θ = 4 cos θ = 1 4 Check that the MODE is in radians. Now use the inverse cosine function. (14) cos 1 ( 1 4) θ k Since 1.57 and 3.14, is between these two numbers, thus θ is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine. See Figure. (15)

7 OpenStax-CNX module: m Figure So, we also need to nd the measure of the angle in quadrant III. In quadrant III, the reference angle is θ' The other solution in quadrant III is θ' The solutions are θ ± k and θ ± k. Try It: Exercise 1 ( on p. 4.) Solve cos θ = Solving Trigonometric Equations in Quadratic Form Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric

8 OpenStax-CNX module: m function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as x or u. If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations. Example 10 Solving a Trigonometric Equation in Quadratic Form Solve the equation exactly: cos θ + 3 cos θ 1 = 0, 0 θ <. We begin by using substitution and replacing cosθ with x. It is not necessary to use substitution, but it may make the problem easier to solve visually. Let cos θ = x. We have x + 3x 1 = 0 (16) The equation cannot be factored, so we will use the quadratic formula x = b± b 4ac a. Replace x with cos θ, and solve. Thus, x = 3± ( 3) 4(1)( 1) = 3± 13 (17) cos θ = 3± 13 ( θ = cos 1 3+ ) 13 ( Note that only the + sign is used. This is because we get an error when we solve θ = cos a calculator, since the domain of the inverse cosine function is [ 1, 1]. However, there is a second solution: ( θ = cos 1 3+ ) This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is ( θ = cos 1 3+ ) (18) ) on (19) (0)

9 OpenStax-CNX module: m Example 11 Solving a Trigonometric Equation in Quadratic Form by Factoring Solve the equation exactly: sin θ 5 sin θ + 3 = 0, 0 θ. Using grouping, this quadratic can be factored. Either make the real substitution, sin θ = u, or imagine it, as we factor: sin θ 5 sin θ + 3 = 0 ( sin θ 3) (sin θ 1) = 0 (1) Now set each factor equal to zero. sin θ 3 = 0 sin θ = 3 sin θ = 3 () sin θ 1 = 0 sin θ = 1 Next solve for θ : sin θ 3, as the range of the sine function is [ 1, 1]. However, sin θ = 1, giving the solution θ =. Analysis Make sure to check all solutions on the given domain as some factors have no solution. Try It: Exercise 15 ( on p. 4.) Solve sin θ = cos θ +, 0 θ. [Hint: Make a substitution to express the equation only in terms of cosine.] Example 1 Solving a Trigonometric Equation Using Algebra Solve exactly: sin θ + sin θ = 0; 0 θ < (3) This problem should appear familiar as it is similar to a quadratic. Let sin θ = x. The equation becomes x + x = 0. We begin by factoring: x + x = 0 x (x + 1) = 0 (4)

10 OpenStax-CNX module: m Set each factor equal to zero. x = 0 (x + 1) = 0 (5) x = 1 Then, substitute back into the equation the original expression sinθ for x. Thus, sin θ = 0 θ = 0, (6) sin θ = 1 θ = 7 6, 11 6 The solutions within the domain 0 θ < are θ = 0,, 7 6, If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero. sin θ + sin θ = 0 sin θ (sin θ + 1) = 0 sin θ = 0 θ = 0, sin θ + 1 = 0 sin θ = 1 (7) sin θ = 1 θ = 7 6, 11 6 Analysis We can see the solutions on the graph in Figure 3. On the interval 0 θ <, the graph crosses the x-axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value.

11 OpenStax-CNX module: m Figure 3 We can verify the solutions on the unit circle in as well. Example 13 Solving a Trigonometric Equation Quadratic in Form Solve the equation quadratic in form exactly: sin θ 3 sin θ + 1 = 0, 0 θ <. We can factor using grouping. values of θ can be found on the unit circle: ( sin θ 1) (sin θ 1) = 0 sin θ 1 = 0 sin θ = 1 θ = 6, 5 6 (8) sin θ = 1 θ = Try It: Exercise 18 ( on p. 4.) Solve the quadratic equation cos θ + cos θ = 0.

12 OpenStax-CNX module: m Solving Trigonometric Equations Using Fundamental Identities While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify the equation. Example 14 Use Identities to Solve an Equation Use identities to solve exactly the trigonometric equation over the interval 0 x <. 3 cos x cos (x) + sin x sin (x) = Notice that the left side of the equation is the dierence formula for cosine. (9) cos x cos (x) + sin x sin (x) = 3 cos (x x) = 3 Dierence formula for cosine cos ( x) = 3 Use the negative angle identity. cos x = 3 From the unit circle in, we see that cos x = 3 when x = 6, (30) Example 15 Solving the Equation Using a Double-Angle Formula Solve the equation exactly using a double-angle formula: cos ( θ) = cos θ. We have three choices of expressions to substitute for the double-angle of cosine. As it is simpler to solve for one trigonometric function at a time, we will choose the double-angle identity involving only cosine: cos (θ) = cos θ cos θ 1 = cos θ cos θ cos θ 1 = 0 ( cos θ + 1) (cos θ 1) = 0 cos θ + 1 = 0 (31) cos θ = 1 cos θ 1 = 0 cos θ = 1 So, if cos θ = 1, then θ = 3 ± k and θ = 4 3 ± k; if cos θ = 1, then θ = 0 ± k.

13 OpenStax-CNX module: m Example 16 Solving an Equation Using an Identity Solve the equation exactly using an identity: 3 cos θ + 3 = sin θ, 0 θ <. If we rewrite the right side, we can write the equation in terms of cosine: 3cos θ + 3 = sin θ 3cos θ + 3 = ( 1 cos θ ) 3cos θ + 3 = cos θ cos θ + 3cos θ + 1 = 0 (cos θ + 1) (cos θ + 1) = 0 cos θ + 1 = 0 (3) cos θ = 1 θ = 3, 4 3 cos θ + 1 = 0 cos θ θ = 1 = Our solutions are θ = 3, 4 3,. 6 Solving Trigonometric Equations with Multiple Angles Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as sin (x) or cos (3x). When confronted with these equations, recall that y = sin (x) is a horizontal compression by a factor of of the function y = sin x. On an interval of, we can graph two periods of y = sin (x), as opposed to one cycle of y = sin x. This compression of the graph leads us to believe there may be twice as many x-intercepts or solutions to sin (x) = 0 compared to sin x = 0. This information will help us solve the equation. Example 17 Solving a Multiple Angle Trigonometric Equation Solve exactly: cos (x) = 1 on [0, ). We can see that this equation is the standard equation with a multiple of an angle. If cos (α) = 1, we know α is in quadrants I and IV. While θ = cos 1 1 will only yield solutions in quadrants I and II, we recognize that the solutions to the equation cos θ = 1 will be in quadrants I and IV. Therefore, the possible angles are θ = 3 and θ = 5 3. So, x = 3 or x = 5 3, which means that x = 6 or x = 5 6. Does this make sense? Yes, because cos ( ( )) ( 6 = cos ) 3 = 1. Are there any other possible answers? Let us return to our rst step. In quadrant I, x = 3, so x = 6 as noted. Let us revolve around the circle again:

14 OpenStax-CNX module: m x = 3 + = (33) = 7 3 so x = 7 6. One more rotation yields x = = = 13 3 x = 13 6 >, so this value for x is larger than, so it is not a solution on [0, ). In quadrant IV, x = 5 3,so x = 5 6 as noted. Let us revolve around the circle again: (34) so x = One more rotation yields x = = = 11 3 (35) x = = = 17 3 x = 17 6 >,so this value for x is larger than,so it is not a solution on [0, ). Our solutions are x = 6, 5 6, , and 6. Note that whenever we solve a problem in the form of sin (nx) = c, we must go around the unit circle n times. (36) 7 Solving Right Triangle Problems We can now use all of the methods we have learned to solve problems that involve applying the properties of right triangles and the Pythagorean Theorem. We begin with the familiar Pythagorean Theorem, a +b = c, and model an equation to t a situation. Example 18 Using the Pythagorean Theorem to Model an Equation Use the Pythagorean Theorem, and the properties of right triangles to model an equation that ts the problem. One of the cables that anchors the center of the London Eye Ferris wheel to the ground must be replaced. The center of the Ferris wheel is 69.5 meters above the ground, and the second anchor

15 OpenStax-CNX module: m on the ground is 3 meters from the base of the Ferris wheel. Approximately how long is the cable, and what is the angle of elevation (from ground up to the center of the Ferris wheel)? See Figure 4. Figure 4 Using the information given, we can draw a right triangle. We can nd the length of the cable with the Pythagorean Theorem. a + b = c (3) + (69.5) m (37) The angle of elevation is θ, formed by the second anchor on the ground and the cable reaching to the center of the wheel. We can use the tangent function to nd its measure. Round to two decimal places.

16 OpenStax-CNX module: m tan θ = tan ( ) The angle of elevation is approximately 71.7, and the length of the cable is 73. meters. (38) Example 19 Using the Pythagorean Theorem to Model an Abstract Problem OSHA safety regulations require that the base of a ladder be placed 1 foot from the wall for every 4 feet of ladder length. Find the angle that a ladder of any length forms with the ground and the height at which the ladder touches the wall. For any length of ladder, the base needs to be a distance from the wall equal to one fourth of the ladder's length. Equivalently, if the base of the ladder is a feet from the wall, the length of the ladder will be 4a feet. See Figure 5. Figure 5 The side adjacent to θ is a and the hypotenuse is 4a. Thus, cos θ = a 4a = 1 4 cos ( ) (39)

17 OpenStax-CNX module: m The elevation of the ladder forms an angle of 75.5 with the ground. The height at which the ladder touches the wall can be found using the Pythagorean Theorem: a + b = (4a) b = (4a) a b = 16a a b = 15a b = 15a (40) Thus, the ladder touches the wall at 15a feet from the ground. Media: Access these online resources for additional instruction and practice with solving trigonometric equations. Solving Trigonometric Equations I 1 Solving Trigonometric Equations II Solving Trigonometric Equations III 3 Solving Trigonometric Equations IV 4 Solving Trigonometric Equations V 5 Solving Trigonometric Equations VI 6 8 Key Concepts When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the dierence of squares, quadratic form, or an expression that lends itself well to substitution. See Example 1, Example, and Example 3. Equations involving a single trigonometric function can be solved or veried using the unit circle. See Example 4, Example 5, and Example 6, and Example 7. We can also solve trigonometric equations using a graphing calculator. See Example 8 and Example 9. Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc. See Example 10, Example 11, Example 1, and Example 13. We can also use the identities to solve trigonometric equation. See Example 14, Example 15, and Example 16. We can use substitution to solve a multiple-angle trigonometric equation, which is a compression of a standard trigonometric function. We will need to take the compression into account and verify that we have found all solutions on the given interval. See Example 17. Real-world scenarios can be modeled and solved using the Pythagorean Theorem and trigonometric functions. See Example

18 OpenStax-CNX module: m Section Exercises 9.1 Verbal Exercise 5 ( on p. 4.) Will there always be solutions to trigonometric function equations? If not, describe an equation that would not have a solution. Explain why or why not. Exercise 6 When solving a trigonometric equation involving more than one trig function, do we always want to try to rewrite the equation so it is expressed in terms of one trigonometric function? Why or why not? Exercise 7 ( on p. 4.) When solving linear trig equations in terms of only sine or cosine, how do we know whether there will be solutions? 9. Algebraic For the following exercises, nd all solutions exactly on the interval 0 θ <. Exercise 8 sin θ = Exercise 9 sin θ = 3 ( on p. 4.) Exercise 30 cos θ = 1 Exercise 31 cos θ = ( on p. 4.) Exercise 3 tan θ = 1 Exercise 33 ( on p. 4.) tan x = 1 Exercise 34 cot x + 1 = 0 Exercise 35 ( on p. 4.) 4 sin x = 0 Exercise 36 csc x 4 = 0 For the following exercises, solve exactly on [0, ). Exercise 37 cos θ = ( on p. 4.) Exercise 38 cos θ = 1 Exercise 39 ( on p. 4.) sin θ = 1 Exercise 40 sin θ = 3 Exercise 41 ( on p. 4.) sin (3θ) = 1

19 OpenStax-CNX module: m Exercise 4 sin (θ) = 3 Exercise 43 cos (3θ) = ( on p. 4.) Exercise 44 cos (θ) = 3 Exercise 45 ( on p. 4.) sin (θ) = 1 Exercise 46 cos ( 5 θ) = 3 For the following exercises, nd all exact solutions on [0, ). Exercise 47 ( on p. 4.) sec (x) sin (x) sin (x) = 0 Exercise 48 tan (x) sin (x) tan (x) = 0 Exercise 49 ( on p. 4.) cos t + cos (t) = 1 Exercise 50 tan (t) = 3 sec (t) Exercise 51 ( on p. 4.) sin (x) cos (x) sin (x) + cos (x) 1 = 0 Exercise 5 cos θ = 1 Exercise 53 ( on p. 4.) sec x = 1 Exercise 54 tan (x) = 1 + tan ( x) Exercise 55 ( on p. 4.) 8 sin (x) + 6 sin (x) + 1 = 0 Exercise 56 tan 5 (x) = tan (x) For the following exercises, solve with the methods shown in this section exactly on the interval [0, ). Exercise 57 ( on p. 4.) sin (3x) cos (6x) cos (3x) sin (6x) = 0.9 Exercise 58 sin (6x) cos (11x) cos (6x) sin (11x) = 0.1 Exercise 59 ( on p. 5.) cos (x) cos x + sin (x) sin x = 1 Exercise 60 6 sin (t) + 9 sin t = 0 Exercise 61 ( on p. 5.) 9 cos (θ) = 9 cos θ 4 Exercise 6 sin (t) = cos t

20 OpenStax-CNX module: m Exercise 63 ( on p. 5.) cos (t) = sin t Exercise 64 cos (6x) cos (3x) = 0 For the following exercises, solve exactly on the interval [0, ). Use the quadratic formula if the equations do not factor. Exercise 65 tan x 3 tan x = 0 ( on p. 5.) Exercise 66 sin x + sin x = 0 Exercise 67 ( on p. 5.) sin x sin x 4 = 0 Exercise 68 5 cos x + 3 cos x 1 = 0 Exercise 69 ( on p. 5.) 3 cos x cos x = 0 Exercise 70 5 sin x + sin x 1 = 0 Exercise 71 ( on p. 5.) tan x + 5tan x 1 = 0 Exercise 7 cot x = cot x Exercise 73 ( on p. 5.) tan x tan x = 0 For the following exercises, nd exact solutions on the interval [0, ). Look for opportunities to use trigonometric identities. Exercise 74 sin x cos x sin x = 0 Exercise 75 ( on p. 5.) sin x + cos x = 0 Exercise 76 sin (x) sin x = 0 Exercise 77 ( on p. 5.) cos (x) cos x = 0 Exercise 78 tan x sec x sin x = cos x Exercise 79 ( on p. 5.) 1 cos (x) = 1 + cos (x) Exercise 80 sec x = 7 Exercise 81 ( on p. 5.) 10 sin x cos x = 6 cos x Exercise 8 3 sin t = 15 cos t sin t

21 OpenStax-CNX module: m Exercise 83 ( on p. 5.) 4 cos x 4 = 15 cos x Exercise 84 8 sin x + 6 sin x + 1 = 0 Exercise 85 ( on p. 5.) 8 cos θ = 3 cos θ Exercise 86 6 cos x + 7 sin x 8 = 0 Exercise 87 ( on p. 5.) 1 sin t + cos t 6 = 0 Exercise 88 tan x = 3 sin x Exercise 89 ( on p. 5.) cos 3 t = cos t 9.3 Graphical For the following exercises, algebraically determine all solutions of the trigonometric equation exactly, then verify the results by graphing the equation and nding the zeros. Exercise 90 6 sin x 5 sin x + 1 = 0 Exercise 91 ( on p. 5.) 8 cos x cos x 1 = 0 Exercise tan x + 0 tan x 3 = 0 Exercise 93 ( on p. 5.) cos x cos x + 15 = 0 Exercise 94 0 sin x 7 sin x + 7 = 0 Exercise 95 ( on p. 5.) tan x + 7 tan x + 6 = 0 Exercise tan x + 69 tan x 130 = Technology For the following exercises, use a calculator to nd all solutions to four decimal places. Exercise 97 ( on p. 5.) sin x = 0.7 Exercise 98 sin x = 0.55 Exercise 99 ( on p. 5.) tan x = 0.34 Exercise 100 cos x = 0.71

22 OpenStax-CNX module: m49398 For the following exercises, solve the equations algebraically, and then use a calculator to nd the values on the interval [0, ). Round to four decimal places. Exercise 101 ( on p. 5.) tan x + 3 tan x 3 = 0 Exercise 10 6 tan x + 13 tan x = 6 Exercise 103 ( on p. 5.) tan x sec x = 1 Exercise 104 sin x cos x = 0 Exercise 105 ( on p. 5.) tan x + 9 tan x 6 = 0 Exercise sin x + sin (x) sec x 3 = Extensions For the following exercises, nd all solutions exactly to the equations on the interval [0, ). Exercise 107 ( on p. 6.) csc x 3 csc x 4 = 0 Exercise 108 sin x cos x 1 = 0 Exercise 109 sin x ( 1 sin x ) + cos x ( 1 sin x ) = 0 ( on p. 6.) Exercise sec x + + sin x tan x + cos x = 0 Exercise 111 ( on p. 6.) sin x 1 + cos (x) cos x = 1 Exercise 11 tan x 1 sec 3 x cos x = 0 Exercise 113 ( on p. 6.) sin(x) sec x = 0 Exercise 114 sin(x) csc x = 0 Exercise 115 ( on p. 6.) cos x sin x cos x 5 = 0 Exercise sec x + + sin x + 4 cos x = 4

23 OpenStax-CNX module: m Real-World Applications Exercise 117 ( on p. 6.) An airplane has only enough gas to y to a city 00 miles northeast of its current location. If the pilot knows that the city is 5 miles north, how many degrees north of east should the airplane y? Exercise 118 If a loading ramp is placed next to a truck, at a height of 4 feet, and the ramp is 15 feet long, what angle does the ramp make with the ground? Exercise 119 ( on p. 6.) If a loading ramp is placed next to a truck, at a height of feet, and the ramp is 0 feet long, what angle does the ramp make with the ground? Exercise 10 A woman is watching a launched rocket currently 11 miles in altitude. If she is standing 4 miles from the launch pad, at what angle is she looking up from horizontal? Exercise 11 ( on p. 6.) An astronaut is in a launched rocket currently 15 miles in altitude. If a man is standing miles from the launch pad, at what angle is she looking down at him from horizontal? (Hint: this is called the angle of depression.) Exercise 1 A woman is standing 8 meters away from a 10-meter tall building. At what angle is she looking to the top of the building? Exercise 13 ( on p. 6.) A man is standing 10 meters away from a 6-meter tall building. Someone at the top of the building is looking down at him. At what angle is the person looking at him? Exercise 14 A 0-foot tall building has a shadow that is 55 feet long. What is the angle of elevation of the sun? Exercise 15 ( on p. 6.) A 90-foot tall building has a shadow that is feet long. What is the angle of elevation of the sun? Exercise 16 A spotlight on the ground 3 meters from a -meter tall man casts a 6 meter shadow on a wall 6 meters from the man. At what angle is the light? Exercise 17 ( on p. 6.) A spotlight on the ground 3 feet from a 5-foot tall woman casts a 15-foot tall shadow on a wall 6 feet from the woman. At what angle is the light? For the following exercises, nd a solution to the following word problem algebraically. Then use a calculator to verify the result. Round the answer to the nearest tenth of a degree. Exercise 18 A person does a handstand with his feet touching a wall and his hands 1.5 feet away from the wall. If the person is 6 feet tall, what angle do his feet make with the wall? Exercise 19 ( on p. 6.) A person does a handstand with her feet touching a wall and her hands 3 feet away from the wall. If the person is 5 feet tall, what angle do her feet make with the wall? Exercise 130 A 3-foot ladder is positioned next to a house. If the ladder slips at 7 feet from the house when there is not enough traction, what angle should the ladder make with the ground to avoid slipping?

24 OpenStax-CNX module: m s to Exercises in this Module to Exercise (p. 3) x = 7 6, 11 6 to Exercise (p. 5) 3 ± k to Exercise (p. 7) θ 1.77 ± k and θ ± k to Exercise (p. 9) cos θ = 1, θ = to Exercise (p. 11), 3, 4 3, 3 to Exercise (p. 18) There will not always be solutions to trigonometric function equations. For a basic example, cos (x) = 5. to Exercise (p. 18) If the sine or cosine function has a coecient of one, isolate the term on one side of the equals sign. If the number it is set equal to has an absolute value less than or equal to one, the equation has solutions, otherwise it does not. If the sine or cosine does not have a coecient equal to one, still isolate the term but then divide both sides of the equation by the leading coecient. Then, if the number it is set equal to has an absolute value greater than one, the equation has no solution. to Exercise (p. 18) 3, 3 to Exercise (p. 18) 3 4, 5 4 to Exercise (p. 18) 4, 5 4 to Exercise (p. 18) 4, 3 4, 5 4, 7 4 to Exercise (p. 18) 4, 7 4 to Exercise (p. 18) 7 6, 11 6 to Exercise (p. 18) 18, 5 18, 13 18, 17 18, 5 18, 9 18 to Exercise (p. 19) 3 1, 5 1, 11 1, 13 1, 19 1, 1 1 to Exercise (p. 19) 1 6, 5 6, 13 6, 17 6, 5 6, 9 6, 37 6 to Exercise (p. 19) 0, 3,, 5 3 to Exercise (p. 19) 3,, 5 3 to Exercise (p. 19) 3, 3, 5 3 to Exercise (p. 19) 0, to Exercise (p. 19) sin 1 ( 1 4), 7 6, 11 6, + sin 1 ( 1 4 )

25 OpenStax-CNX module: m to Exercise (p. 19) ( ( )) 1 9 sin 1, ( sin 1 ( 9 10 to Exercise (p. 19) 0 to Exercise (p. 19) 6, 5 6, 7 6, 11 6 to Exercise (p. 19) 3, 6, 5 6 to Exercise (p. 0) 0, 3,, 4 3 to Exercise (p. 0) There are no solutions. to Exercise (p. 0) cos ( ( )) ( , cos ( to Exercise )) (p. 0) ( 9 5, + tan 1 1 tan ( 1 1 to Exercise (p. 0) There are no solutions. to Exercise (p. 0) There are no solutions. to Exercise (p. 0) 0, 3, 4 3 to Exercise (p. 0) 4, 3 4, 5 4, 7 4 ) to Exercise (p. 0), sin 1 ( 3 5, ( ) sin 1 3 5, 3 to Exercise (p. 0) cos ( 1 4) ( ) 1, cos to Exercise (p. 1) )), ( ( )) sin 1, 1 ( ( )) 9 sin 1, 4 ( ( )) sin 1, 5 ( sin 1 3 ( 1 7 )) 3, ( cos 1 4) ( ) 3, cos 1 3 4, 5 3 to Exercise (p. 1) cos ( 1 3 4) (, cos 1 3) (, cos 1 3 to Exercise (p. 1) 0,,, 3 to Exercise (p. 1) ), 5 3 ( )) ( ( )) ( ( )) 9 5, + tan , + tan ) ( ), cos , ( cos 1 4) ( 1, cos to Exercise (p. 1) There are no solutions. to Exercise (p. 1) + tan 1 ( ), + tan ( 1 ) 3, + tan 1 ( ), + tan ( 1 3 to Exercise (p. 1) k , k to Exercise (p. 1) k to Exercise (p. ) , 1.887, , to Exercise (p. ) 1.047, , )

26 OpenStax-CNX module: m to Exercise (p. ) 0.536, , 3.674, to Exercise (p. ) sin 1 ( ) 1 4, sin 1 ( ) 1 4, 3 to Exercise (p. ), 3 to Exercise (p. ) There are no solutions. to Exercise (p. ) 0,,, 3 to Exercise (p. ) There are no solutions. to Exercise (p. 3) 7. to Exercise (p. 3) 5.7 to Exercise (p. 3) 8.4 to Exercise (p. 3) 31.0 to Exercise (p. 3) 88.7 to Exercise (p. 3) 59.0 to Exercise (p. 3) 36.9

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