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Pre-Class Problems 16 for Tuesday, November 18 These are the type of problems that you will be working on in class. These problems are from Lesson 10. Solution to Problems on the Pre-Exam. You can go to the solution for each problem by clicking on the problem letter. Objective of the following problems: To find all the solutions to a trigonometric equation. 1. Find all the exact solutions for the following equations. a. cos 0 b. 5 sin 8 3 c. 6 sec 5 1 d. sin x 0 e. tan 4 f. 8 sec 15 3 Objective of the following problems: To find all the solutions to a trigonometric equation in a specified interval. 3 2a. Find all the exact solutions (in radians) of the equation cos 3 x in 2 the interval [ , ] . Put a box around your answer(s). 2b. Find all the exact solutions (in degrees) of the equation 2 sin in the interval [ 900 , 1620 ] . 2c. Find all the exact solutions (in radians) 3 cot ( 2 ) 3 0 in the interval [ 0 , 2 ] . of 2d. Find all the exact solutions (in degrees) of the equation 5 sec ( 4 80 ) 6 4 in the interval [ 540 , 270 ] . 2 2 0 the equation 2e. Find all the exact solutions (in radians) of 3 2 sin x 3 0 in the interval [ 3 , 5 ] . 4 2f. Find all the exact solutions (in degrees) of the equation sin 6 0 in the interval [ 90 , 180 ] . 2g. Find all the exact solutions (in radians) of 2 7 sec 5 11 4 in the interval [ , 2 ] . 3 the the equation equation Additional problems available in the textbook: Page 231 … 39 - 44, 45. Examples 7 and 8 on page 229. Solutions: 1a. cos 0 Back to Problem 1. NOTE: The angle solutions for this equation must lie on one of the coordinate axes since zero is not positive nor negative. NOTE: Using Unit Circle Trigonometry, we know that cos the one solution, which lies on the positive y-axis, is All solutions on the positive y-axis: . 2 0 . Thus, 2 2 n , where n is an integer 2 3 0 . Thus, 2 3 the one solution, which lies on the negative y-axis, is . Also, using 2 NOTE: Using Unit Circle Trigonometry, we know that cos Unit Circle Trigonometry, we know that cos 0 . 2 solution, which lies on the negative y-axis, is All solutions on the negative y-axis: Thus, the one . 2 3 2 n or 2 n , 2 2 where n is an integer Answer: 3 2 n , 2 n , where n is an integer 2 2 OR 2 n , 2 n , where n is an integer 2 2 3 2 n , 2 n , where n is an 2 2 n , where n is an integer. integer, may also be written as 2 NOTE: 1b. The answer 5 sin 8 3 Back to Problem 1. NOTE: In order to solve this equation, we must first isolate sin on one side of the equation by itself. 5 sin 8 3 5 sin 5 sin 1 NOTE: The angle solutions for this equation must lie on one of the coordinate axes since 1 is the minimum negative number in the range of the sine function. 3 1 . Thus, 2 3 the one solution, which lies on the negative y-axis, is . Also, using 2 Unit Circle Trigonometry, we know that sin 1 . Thus, the one 2 solution, which lies on the negative y-axis, is . 2 NOTE: Using Unit Circle Trigonometry, we know that sin All solutions on the negative y-axis: 3 2 n or 2 n , 2 2 where n is an integer Answer: 1c. 3 2 n or 2 n , where n is an integer 2 2 6 sec 5 1 Back to Problem 1. NOTE: In order to solve this equation, we must first isolate sec on one side of the equation by itself. 6 sec 5 1 6 sec 6 sec 1 NOTE: Now, we will take the reciprocal of both sides in order to obtain cos on the left side of the equation. sec 1 cos 1 NOTE: The angle solutions for this equation must lie on one of the coordinate axes since 1 is the maximum positive number in the range of the cosine function. NOTE: Using Unit Circle Trigonometry, we know that cos 0 1 . Thus, the one solution, which lies on the positive x-axis, is 0 . All solutions on the positive x-axis: 0 2 n 2 n , where n is an integer Answer: 2 n , where n is an integer 1d. sin x 0 Back to Problem 1. NOTE: The angle solutions for this equation must lie on one of the coordinate axes since zero is not positive nor negative. NOTE: Using Unit Circle Trigonometry, we know that sin 0 0 . Thus, the one solution, which lies on the positive x-axis, is x 0 . All solutions on the positive x-axis: x 0 2 n 2 n , where n is an integer NOTE: Using Unit Circle Trigonometry, we know that sin 0 . Thus, the one solution, which lies on the negative x-axis, is x . All solutions on the negative x-axis: x 2 n ( 2 n 1 ) , where n is an integer Answer: x 2 n , x ( 2 n 1 ) , where n is an integer NOTE: This answer may also be written as x n , where n is an integer. 1e. tan 4 Back to Problem 1. NOTE: Tangent is negative in the second and fourth quadrants. If we find one angle solution in the second quadrant, then all the other angle solutions in the second quadrant are coterminal to this one. The same is true for all the solutions in the fourth quadrant. To find the reference angle ' : tan 4 tan ' 4 ' tan 1 4 NOTE: The one solution in the second quadrant that is in the interval [ 0 , 2 ) is given by ' tan 1 4 . All solutions in II: where n is an integer tan 1 4 2 n tan 1 4 ( 2 n 1) , NOTE: The one solution in the fourth quadrant that is in the interval [ 0 , 2 ) 1 is given by 2 ' 2 tan 4 . The one solution in the fourth 1 quadrant that is in the interval ( 2 , 0 ] is given by ' tan 4 . 1 1 All solutions in IV: 2 tan 4 2 n tan 4 2 ( n 1) 1 or tan 4 2 n , where n is an integer Answer: tan 1 4 ( 2 n 1) , tan 1 4 2 ( n 1) , where n is an integer OR tan 1 4 ( 2 n 1) , tan 1 4 2 n , where n is an integer 1f. 8 sec 15 3 Back to Problem 1. NOTE: In order to solve this equation, we must first isolate sec on one side of the equation by itself. 8 sec 15 3 8 sec 12 sec 12 3 sec 8 2 NOTE: Now, we will take the reciprocal of both sides in order to obtain cos on the left side of the equation. sec 3 2 cos 2 3 NOTE: Cosine is positive in the first and fourth quadrants. If we find one angle solution in the first quadrant, then all the other angle solutions in the first quadrant are coterminal to this one. The same is true for all the solutions in the fourth quadrant. To find the reference angle ' : cos 2 2 2 cos ' ' cos 1 3 3 3 NOTE: The one solution in the first quadrant that is in the interval [ 0 , 2 ) 1 2 ' cos is given by . 3 1 All solutions in I: cos 2 2 n , where n is an integer 3 NOTE: The one solution in the fourth quadrant that is in the interval 1 2 ( 2 , 0 ] is given by ' cos . 3 1 All solutions in IV: cos 1 Answer: cos 2a. cos 3 x 2 2 n , where n is an integer 3 2 2 2 n , cos 1 2 n , where n is an integer 3 3 3 , [ , ] 2 Back to Problem 2. Let u 3 x . Then cos 3 x 3 3 cos u 2 2 NOTE: Cosine is negative in the second and third quadrants. To find the reference angle u ' : cos u 3 cos u ' 2 3 3 u ' cos 1 2 2 6 NOTE: The one solution in the second quadrant that is in the interval 6 5 [ 0 , 2 ) is given by u u ' . 6 6 6 6 All solutions for u in II: u 5 2 n , where n is an integer 6 5 2 n and u 3 x 6 5 5 2 n 3x 2 n x , where n is an integer. 6 18 3 All solutions for x in II: u Simplifying, we have that x = 5 2 n 5 12 n 5 12 n = = 18 3 18 18 18 (12 n 5 ) , where n is an integer 18 The solutions for x n 0: x ( 12 n 5 ) which are in the interval [ , ] : 18 ( 0 5) 18 5 18 n 1: x ( 12 5 ) 18 n 2: x ( 24 5 ) 29 18 18 NOTE: 17 18 29 18 . So, this solution is not in the interval [ , ] . 18 18 n 1: x ( 12 5 ) 7 18 18 n 2: x ( 24 5 ) 19 18 18 NOTE: [ , ]. 19 18 . So, this solution is not in the interval 18 18 NOTE: The one solution in the third quadrant that is in the interval [ 0 , 2 ) 6 7 is given by u u ' . 6 6 6 6 All solutions for u in III: u 7 2 n , where n is an integer 6 7 2 n and u 3 x 6 7 7 2 n 3x 2 n x , where n is an integer. 6 18 3 All solutions for x in III: u Simplifying, we have that x = 7 2 n 7 12 n 7 12 n = = 18 3 18 18 18 ( 12 n 7 ) , where n is an integer 18 The solutions for x ( 12 n 7 ) which are in the interval [ , ] : 18 n 0: x ( 0 7 ) 18 7 18 n 1: x ( 12 7 ) 18 19 18 NOTE: 19 18 . So, this solution is not in the interval [ , ] . 18 18 n 1: x ( 12 7 ) 5 18 18 n 2: x ( 24 7 ) 17 18 18 n 3: x ( 36 7 ) 29 29 18 18 18 NOTE: [ , ]. Answer: 5 18 17 18 29 18 . So, this solution is not in the interval 18 18 5 7 17 , , 18 18 18 2b. 2 sin 2 2 0 , [ 900 , 1620 ] Let u . Then 2 sin 2 2 2 sin u 2 sin u 2 0 2 sin u Back to Problem 2. 2 0 2 2 NOTE: Sine is positive in the first and second quadrants. To find the reference angle u ' : sin u 2 sin u ' 2 2 2 u ' sin 1 45 2 2 NOTE: The one solution in the first quadrant that is in the interval [ 0 , 360 ) is given by u u ' 45 . All solutions for u in I: u 45 n 360 , where n is an integer All solutions for in I: u 45 n 360 and u 2 45 n 360 90 n 720 , where n is an integer. 2 The solutions for 90 n 720 which are in the interval [ 900 , 1620 ] : n 0 : 90 0 n 1 : 90 720 n 2 : 90 1440 90 810 1530 n 3 : 90 2160 2250 NOTE: 2250 1620 . So, this solution is not in the interval [ 900 , 1620 ] . n 1 : 90 720 630 n 2 : 90 1440 1350 1350 900 . NOTE: [ 900 , 1620 ] . So, this solution is not in the interval NOTE: The one solution in the second quadrant that is in the interval [ 0 , 360 ) is given by u 180 u ' 180 45 135 . All solutions for u in II: u 135 n 360 , where n is an integer All solutions for in II: u 135 n 360 and u 2 135 n 360 270 n 720 , where n is an integer. 2 The solutions for 270 n 720 which are in the interval [ 900 , 1620 ] : n 0 : 270 0 270 n 1 : 270 720 990 n 2 : 270 1440 1710 NOTE: 1710 1620 . So, this solution is not in the interval [ 900 , 1620 ] . n 1 : 270 720 450 n 2 : 270 1440 1170 1170 900 . NOTE: [ 900 , 1620 ] . So, this solution is not in the interval Answer: 630 , 450 , 90 , 270 , 810 , 990 , 1530 2c. 3 cot ( 2 ) 3 0 , [ 0 , 2 ] Let u 2 . Then 3 cot ( 2 ) 3 cot u 3 0 3 cot u tan u 3 tan u 3 Back to Problem 2. 3 0 3 cot u 3 3 3 NOTE: Tangent is negative in the second and fourth quadrants. To find the reference angle u ' : tan u 3 tan u ' 3 u ' tan 1 3 3 NOTE: The one solution in the second quadrant that is in the interval 3 2 [ 0 , 2 ) is given by u u ' . 3 3 3 3 All solutions for u in II: u 2 2 n , where n is an integer 3 2 2 n and u 2 3 2 2 2 n 2 2 n n , 3 3 6 where n is an integer. All solutions for in II: u Simplifying, we have that = 6 n 6 n n = = 6 6 6 6 ( 6 n 1) , where n is an integer 6 The solutions for n 0: NOTE: ( 6 n 1) which are in the interval [ 0 , 2 ] : 6 ( 0 1) 6 6 6 0 . So, this solution is not in the interval [ 0 , 2 ] . 6 n 1: ( 6 1) 6 n 2: ( 12 1 ) 6 n 3: (18 1) 17 6 6 NOTE: 5 6 11 6 17 12 2 . So, this solution is not in the interval [ 0 , 2 ] . 6 6 n 1: ( 6 1) 7 6 6 7 6 NOTE: NOTE: 7 0 . So, this solution is not in the interval [ 0 , 2 ] . 6 The one solution in the fourth quadrant that is in the interval ( 2 , 0 ] is given by u u ' All solutions for u in IV: u . 3 2 n , where n is an integer 3 2 n and u 2 3 4 2 2 2 n 2 2 n n , 3 3 3 All solutions for in IV: u where n is an integer. Simplifying, we have that 2 2 3n n = = 3 3 3 2 3 n ( 3n 2 ) = , where n is an integer 3 3 The solutions for n 0: NOTE: ( 3n 2 ) which are in the interval [ 0 , 2 ] : 3 ( 0 2 ) 2 2 3 3 3 2 0 . So, this solution is not in the interval [ 0 , 2 ] . 3 n 1: ( 3 2 ) 3 3 n 2: ( 6 2 ) 3 4 3 n 3: NOTE: ( 9 2 ) 7 3 3 7 6 2 . So, this solution is not in the interval [ 0 , 2 ] . 3 3 n 1: 2d. ( 3 2 ) 5 3 3 5 3 NOTE: 5 0 . So, this solution is not in the interval [ 0 , 2 ] . 3 Answer: 5 4 11 , , , 3 6 3 6 5 sec ( 4 80 ) 6 4 , [ 540 , 270 ] Back to Problem 2. Let u 4 80 . Then 5 sec ( 4 80 ) 6 4 5 sec u 6 4 5 sec u 10 sec u 2 cos u 1 2 NOTE: Cosine is positive in the first and fourth quadrants. To find the reference angle u ' : cos u 1 1 1 cos u ' u ' cos 1 60 2 2 2 NOTE: The one solution in the first quadrant that is in the interval [ 0 , 360 ) is given by u u ' 60 . All solutions for u in I: u 60 n 360 , where n is an integer All solutions for in I: u 60 n 360 and u 4 80 4 80 60 n 360 4 140 n 360 35 n 90 , where n is an integer. The solutions for 35 n 90 which are in the interval [ 540 , 270 ] : n 0 : 35 0 35 n 1 : 35 90 125 n 2 : 35 180 215 n 3 : 35 270 305 NOTE: 305 270 . So, this solution is not in the interval [ 540 , 270 ] . n 1 : 35 90 55 n 2 : 35 180 145 n 3 : 35 270 235 n 4 : 35 360 325 n 5 : 35 450 415 n 6 : 35 540 505 n 7 : 35 630 595 595 540 . NOTE: [ 540 , 270 ] . So, this solution is not in the interval NOTE: The one solution in the fourth quadrant that is in the interval ( 360 , 0 ] is given by u u ' 60 . All solutions for u in IV: u 60 n 360 , where n is an integer All solutions for in IV: u 60 n 360 and u 4 80 4 80 60 n 360 4 20 n 360 5 n 90 , where n is an integer. The solutions for 5 n 90 which are in the interval [ 540 , 270 ] : n 0 : 5 0 5 n 1 : 5 90 95 n 2 : 5 180 185 n 3 : 5 270 275 NOTE: 275 270 . So, this solution is not in the interval [ 540 , 270 ] . n 1 : 5 90 85 n 2 : 5 180 175 n 3 : 5 270 265 n 4 : 5 360 355 n 5 : 5 450 445 n 6 : 5 540 535 n 7 : 5 630 625 625 540 . NOTE: [ 540 , 270 ] . So, this solution is not in the interval Answer: 535 , 505 , 445 , 415 , 355 , 325 , 265 , 235 , 175 , 145 , 85 , 55 , 5 , 35 , 95 , 125 , 185 , 215 2e. 3 2 sin x 4 Let u x 2 sin u 3 0 , [ 3 , 5 ] 3 3 2 sin x . Then 4 4 3 0 2 sin u Back to Problem 2. 3 0 3 sin u 3 2 NOTE: Sine is negative in the third and fourth quadrants. To find the reference angle u ' : sin u 3 sin u ' 2 3 3 u ' sin 1 2 2 3 NOTE: The one solution in the third quadrant that is in the interval [ 0 , 2 ) 3 4 is given by u u ' . 3 3 3 3 4 2 n , where n is an integer 3 All solutions for u in III: u 3 4 2 n and u x 4 3 3 4 7 x 2 n x 2 n , where n is an integer. 4 3 12 All solutions for x in III: u NOTE: 4 3 16 9 7 3 4 12 12 12 7 7 24 n 2 n = = 12 12 12 Simplifying, we have that x 7 24 n ( 24 n 7 ) = , where n is an integer 12 12 The solutions for x ( 24 n 7 ) which are in the interval [ 3 , 5 ] : 12 n 0: x ( 0 7 ) 12 7 12 n 1: x ( 24 7 ) 12 31 12 n 2: x ( 48 7 ) 12 55 12 n 3: x ( 72 7 ) 79 12 12 79 60 5 . 12 12 [ 3 , 5 ] . NOTE: So, this solution is not in the interval n 1: x ( 24 7 ) 17 12 12 17 12 n 2: x ( 48 7 ) 41 12 12 41 12 41 36 3 . So, this solution is not in the interval 12 12 [ 3 , 5 ] . NOTE: NOTE: The one solution in the fourth quadrant that is in the interval ( 2 , 0 ] is given by u u ' All solutions for u in IV: u . 3 2 n , where n is an integer 3 3 2 n and u x 4 3 3 13 x 2 n x 2 n , where n is an integer. 4 3 12 All solutions for x in IV: u NOTE: 3 4 9 13 3 4 12 12 12 13 13 24 n 2 n = = 12 12 12 13 24 n ( 24 n 13 ) = , where n is an integer 12 12 Simplifying, we have that x The solutions for x n 0: x ( 24 n 13 ) which are in the interval [ 3 , 5 ] : 12 ( 0 13 ) 13 12 12 13 12 n 1: x ( 24 13 ) 12 11 12 n 2: x ( 48 13 ) 12 35 12 n 3: x ( 72 13 ) 12 59 12 n 4: x ( 96 13 ) 83 12 12 83 60 5 . 12 12 [ 3 , 5 ] . NOTE: So, this solution is not in the interval n 1: x ( 24 13 ) 37 12 12 37 12 n 2: x ( 48 13 ) 61 12 12 61 12 61 36 3 . So, this solution is not in the interval 12 12 [ 3 , 5 ] . NOTE: Answer: 2f. 37 17 13 7 11 31 35 55 59 , , , , , , , , 12 12 12 12 12 12 12 12 12 sin 6 0 , [ 90 , 180 ] Let u 6 . Then sin 6 0 sin u 0 Back to Problem 2. NOTE: The angle solutions for this equation must lie on one of the coordinate axes since zero is not positive nor negative. NOTE: Using Unit Circle Trigonometry, we know that sin 0 0 . Thus, the one solution, which lies on the positive x-axis, is u 0 . All solutions for u on the positive x-axis: where n is an integer u 0 n 360 n 360 , NOTE: Using Unit Circle Trigonometry, we know that sin 180 0 . Thus, the one solution, which lies on the negative x-axis, is u 180 . All solutions for u on the negative x-axis: u 180 n 360 , where n is an integer NOTE: These two solutions of u 0 n 360 and u 180 n 360 , where n is an integer, may be written as u 0 n 180 n 180 , where n is an integer. All solutions for : u n 180 and u 6 6 n 180 n 30 , where n is an integer. The solutions for n 30 which are in the interval [ 90 , 180 ] : n 0: 0 n 1: 30 n 2: 60 n 3: 90 n 4: 120 n 5: 150 n 6: 180 n 1: 30 n 2: 60 n 3: 90 Answer: 90 , 60 , 30 , 0 , 30 , 60 , 90 , 120 , 150 , 180 2g. 2 7 sec 5 11 4 , [ , 2 ] 3 Let u 5 Back to Problem 2. 2 2 11 4 . Then 7 sec 5 3 3 7 sec u 11 4 7 sec u 7 sec u 1 cos u 1 NOTE: The angle solutions for this equation must lie on one of the coordinate axes since 1 is the minimum negative number in the range of the cosine function. NOTE: Using Unit Circle Trigonometry, we know that cos 1 . Thus, the one solution, which lies on the negative x-axis, is u . All solutions for u on the negative x-axis: u 2 n , where n is an integer 2 3 2 5 2 n 5 2 n 5 2 n , 3 3 3 5 where n is an integer. All solutions for : u 2 n and u 5 Simplifying, we have that 2 n 5 6 n = = 15 15 3 5 5 6 n ( 6 n 5) = , where n is an integer 15 15 The solutions for ( 6 n 5) which are in the interval [ , 2 ] : 15 n 0: ( 0 5) 5 15 15 n 1: ( 6 5) 15 n 2: ( 12 5 ) 15 17 15 n 3: ( 18 5 ) 15 23 15 n 4: ( 24 5 ) 15 29 15 n 5: ( 30 5 ) 35 7 15 15 3 NOTE: 3 11 15 7 6 2 . So, this solution is not in the interval [ , 2 ] . 3 3 n 1: ( 6 5) 15 n 2: ( 12 5 ) 15 15 7 15 n 3: ( 18 5 ) 15 13 15 n 4: ( 24 5 ) 15 19 15 NOTE: [ , 2 ] . Answer: 19 15 . So, this solution is not in the interval 15 15 13 7 11 17 23 29 , , , , , , , 15 15 15 3 15 15 15 15 Solution to Problems on the Pre-Exam: 25. Back to Page 1. 3 in the 2 interval [ 180 , 180 ] . (9 pts.) Put a box around your answer(s). Find all the solutions (in degrees) to the equation cos 3 x Let u 3 x . Then cos 3 x 3 3 cos u 2 2 NOTE: Cosine is negative in the second and third quadrants. To find the reference angle u ' : cos u 3 cos u ' 2 3 3 u ' cos 1 30 2 2 NOTE: The one solution in the second quadrant that is in the interval [ 0 , 360 ) is given by u 180 u ' 180 30 150 . All solutions for u in II: u 150 n 360 , where n is an integer All solutions for x in II: u 150 n 360 and u 3 x 3 x 150 n 360 x 50 n 120 , where n is an integer. The solutions for x 50 n 120 which are in the interval [ 180 , 180 ] : n 0 : x 50 n 1 : x 50 120 170 n 2 : x 50 240 290 180 n 1 : x 50 120 70 n 2 : x 50 240 190 180 NOTE: The one solution in the third quadrant that is in the interval [ 0 , 360 ) is given by u 180 u ' 180 30 210 . All solutions for u in III: u 210 n 360 , where n is an integer All solutions for x in III: u 210 n 360 and u 3 x 3 x 210 n 360 x 70 n 120 , where n is an integer. n 0 : x 70 n 1 : x 70 120 190 180 n 1 : x 70 120 50 n 2 : x 70 240 170 n 3 : x 70 360 290 180