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5-3 Solving Trigonometric Equations Solve each equation for all values of x. 1. 5 sin x + 2 = sin x SOLUTION: The period of sine is 2π, so you only need to find solutions on the interval and . Solutions on the interval (– general form of the solutions is , . The solutions on this interval are ), are found by adding integer multiples of 2π. Therefore, the + 2nπ, + 2nπ, . 3. 2 = 4 cos2 x + 1 SOLUTION: The period of cosine is 2π, so you only need to find solutions on the interval are , , , and . Solutions on the interval (– Therefore, the general form of the solutions is , + 2nπ, . The solutions on this interval ), are found by adding integer multiples of 2π. + 2nπ, + 2nπ, + 2nπ, . 5. 9 + cot2 x = 12 SOLUTION: The period of cotangent is π, so you only need to find solutions on the interval are and . Solutions on the interval (– general form of the solutions is eSolutions Manual - Powered by Cognero 7. 3 csc x = 2 csc x + + nπ, , . The solutions on this interval ), are found by adding integer multiples of π. Therefore, the + nπ, . Page 1 are , , , and . Solutions on the interval (– 5-3 Therefore, Solving Trigonometric Equations the general form of the solutions is , ), are found by adding integer multiples of 2π. + 2nπ, + 2nπ, + 2nπ, + 2nπ, . 5. 9 + cot2 x = 12 SOLUTION: The period of cotangent is π, so you only need to find solutions on the interval are and . Solutions on the interval (– general form of the solutions is + nπ, . The solutions on this interval ), are found by adding integer multiples of π. Therefore, the , + nπ, . 7. 3 csc x = 2 csc x + SOLUTION: The period of cosecant is 2π, so you only need to find solutions on the interval are and . Solutions on the interval (– general form of the solutions is + 2nπ, , . The solutions on this interval ), are found by adding integer multiples of 2π. Therefore, the + 2nπ, . 9. 6 tan2 x – 2 = 4 SOLUTION: The period of tangent is π, so you only need to find solutions on the interval and . Solutions on the interval (– form of the solutions is 11. 7 cot x – + nπ, + nπ, , . The solutions on this interval are ), are found by adding integer multiples of π. Therefore, the general . = 4 cot x SOLUTION: eSolutions Manual - Powered by Cognero The period of cotangent is π, so you only need to find solutions on the interval Page 2 . The only solution on this and . Solutions on the interval (– ), are found by adding integer multiples of π. Therefore, the general , 5-3 form Solving of theTrigonometric solutions is + nπ, Equations + nπ, 11. 7 cot x – . = 4 cot x SOLUTION: The period of cotangent is π, so you only need to find solutions on the interval interval is . Solutions on the interval (– form of the solutions is + nπ, , . The only solution on this ), are found by adding integer multiples of π. Therefore, the general . Find all solutions of each equation on [0, 2 ). 13. sin4 x + 2 sin2 x − 3 = 0 SOLUTION: when x = On the interval [0, 2π), and when x = . Since is not a real number, the yields no additional solutions. equation 15. 4 cot x = cot x sin2 x SOLUTION: The equations sin x = 2 and sin x = –2 have no real solutions. On the interval [0, 2π), the equation cot x = 0 has solutions and . 2 eSolutions - Powered by Cognero 17. cos3Manual x + cos x – cos x=1 SOLUTION: Page 3 The equations sin x = 2 and sin x = –2 have no real solutions. On the interval [0, 2π), the equation cot x = 0 has 5-3 solutions Solving Trigonometric Equations and . 17. cos3 x + cos2 x – cos x = 1 SOLUTION: On the interval [0, 2π), the equation cos x = 1 has a solution of 0 and the equation cos x = –1 has a solution of π. 19. TENNIS A tennis ball leaves a racket and heads toward a net 40 feet away. The height of the net is the same height as the initial height of the tennis ball. a. If the ball is hit at 50 feet per second, neglecting air resistance, use d = 2 v0 sin 2 to find the interval of possible angles of the ball needed to clear the net. b. Find if the initial velocity remained the same but the distance to the net was 50 feet. SOLUTION: a. The interval is [15.4°, 74.6°]. b. eSolutions Manual - Powered by Cognero Page 4 5-3 Solving Trigonometric Equations On the interval [0, 2π), the equation cos x = 1 has a solution of 0 and the equation cos x = –1 has a solution of π. 19. TENNIS A tennis ball leaves a racket and heads toward a net 40 feet away. The height of the net is the same height as the initial height of the tennis ball. a. If the ball is hit at 50 feet per second, neglecting air resistance, use d = 2 v0 sin 2 to find the interval of possible angles of the ball needed to clear the net. b. Find if the initial velocity remained the same but the distance to the net was 50 feet. SOLUTION: a. The interval is [15.4°, 74.6°]. b. If the distance to the net is 50 feet, then the angle would be 19.9° or 70.1°. Find all solutions of each equation on the interval [0, 2 ). 21. 1 = cot 2 x + csc x eSolutions Manual - Powered by Cognero SOLUTION: Page 5 5-3 Solving Trigonometric Equations If the distance to the net is 50 feet, then the angle would be 19.9° or 70.1°. Find all solutions of each equation on the interval [0, 2 ). 21. 1 = cot 2 x + csc x SOLUTION: Therefore, on the interval [0, 2π) the solutions are , , and . 23. tan 2 x = 1 – sec x SOLUTION: Therefore, on the interval [0, 2π) the solutions are 0, , and . 25. 2 – 2 cos2 x = sin x + 1 SOLUTION: eSolutions Manual - Powered by Cognero Page 6 5-3 Therefore, Solving on Trigonometric the interval [0, 2π)Equations the solutions are 0, , and . 25. 2 – 2 cos2 x = sin x + 1 SOLUTION: Therefore, on the interval [0, 2π) the solutions are , , and . 27. 3 sin x = 3 – 3 cos x SOLUTION: Therefore, on the interval [0, 2π) the only valid solutions are 29. sec2 x – 1 + tan x – and 0. tan x = SOLUTION: eSolutions Manual - Powered by Cognero Page 7 5-3 Therefore, Solving on Trigonometric the interval [0, 2π)Equations the only valid solutions are 29. sec2 x – 1 + tan x – and 0. tan x = SOLUTION: Therefore, on the interval [0, 2π) the solutions are , , and . 31. OPTOMETRY Optometrists sometimes join two oblique or tilted prisms to correct vision. The resultant refractive power PR of joining two oblique prisms can be calculated by first resolving each prism into its horizontal and vertical components, PH and PV. Using the equations above, determine for what values of PV and PH are equivalent. SOLUTION: The sine and cosine have the same values in the interval [0, 2π) at eSolutions Manual -the Powered by Cognero Therefore, components will be equivalent when Find all solutions of each equation on the interval [0, 2 ). and . . Page 8 5-3 Therefore, Solving on Trigonometric the interval [0, 2π)Equations the solutions are , , and . 31. OPTOMETRY Optometrists sometimes join two oblique or tilted prisms to correct vision. The resultant refractive power PR of joining two oblique prisms can be calculated by first resolving each prism into its horizontal and vertical components, PH and PV. Using the equations above, determine for what values of PV and PH are equivalent. SOLUTION: and The sine and cosine have the same values in the interval [0, 2π) at Therefore, the components will be equivalent when . . Find all solutions of each equation on the interval [0, 2 ). 33. + = –4 SOLUTION: On the interval [0, 2π),cos x = – eSolutions Manual - Powered by Cognero 35. cot x cos x + 1 = + when x = and when x = . Page 9 and The sine and cosine have the same values in the interval [0, 2π) at 5-3 Therefore, Solving the Trigonometric Equations components will be equivalent when . . Find all solutions of each equation on the interval [0, 2 ). 33. + = –4 SOLUTION: On the interval [0, 2π),cos x = – 35. cot x cos x + 1 = when x = and when x = . + SOLUTION: On eSolutions Manual - Powered by Cognero , Page 10 the interval [0, 2π),cos x = – Equations when x = 5-3 On Solving Trigonometric 35. cot x cos x + 1 = and when x = . + SOLUTION: On when x = and when x = , . GRAPHING CALCULATOR Solve each equation on the interval [0, 2 ) by graphing. Round to the nearest hundredth. 37. sin x + cos x = 3x SOLUTION: On the interval , the only solution is when x = 0.41. 39. x log x + 5x cos x = –2 SOLUTION: eSolutions Manual - Powered by Cognero Page 11 5-3 On Solving Trigonometric the interval , the only Equations solution is when x = 0.41. 39. x log x + 5x cos x = –2 SOLUTION: On the interval , the solutions are when x = 1.84 and when x = 4.49. Find the x-intercepts of each graph on the interval [0, 2 ). 41. SOLUTION: Let y = 0 and solve for x. On the interval [0, 2π) cos x = 0 when x = 43. eSolutions Manual - Powered by Cognero SOLUTION: Let y = 0 and solve for x. and x = . Page 12 the interval [0, 2π) cos x = 0Equations when x = and x = 5-3 On Solving Trigonometric . 43. SOLUTION: Let y = 0 and solve for x. and x = On the interval [0, 2π) cot x = 1 when x = . Find all solutions of each equation on the interval [0, 4 ). 45. 4 tan x = 2 sec2 x SOLUTION: On the interval [0, 4π) the solutions are , , , and . 47. csc x cot2 x = csc x SOLUTION: eSolutions Manual - Powered by Cognero Page 13 the interval [0, 4π) the solutions are , 5-3 On Solving Trigonometric Equations , , and , , . 47. csc x cot2 x = csc x SOLUTION: On the interval [0, 4π) the solutions are , , , , , and . 49. GEOMETRY Consider the circle below. a. The length s of is given by s = r(2 ) where 0 ≤ ≤ . When s = 18 and AB = 14, the radius is r = Use a graphing calculator to find the measure of 2 . in radians. b. The area of the shaded region is given by A = . Use a graphing calculator to find the radian measure of θ if the radius is 5 inches and the area is 36 square inches. Round to the nearest hundredth. SOLUTION: a. Rewrite the arclength formula using s = 18 and r = . On the graphing calculator, find the intersection of Y1 = 18sinθ and Y2 = 14θ. eSolutions Manual - Powered by Cognero The value of 2θ = 2(1.1968) or about 2.39 radians. Page 14 5-3 Solving Trigonometric Equations On the graphing calculator, find the intersection of Y1 = 18sinθ and Y2 = 14θ. The value of 2θ = 2(1.1968) or about 2.39 radians. b. First, substitute into the given area formula and rearrange it. Using a graphing calculator, find the intersection of Y1 = 2.88 and Y2 = θ – sin θ. When the area is 36 square inches and the radius is 5 inches, then the measure of θ is 3.01 radians. Solve each inequality on the interval [0, 2 ). 51. 0 < 2 cos x – SOLUTION: Graph y = 2 cos x – < 2 cos x – . on [0, 2π). Use the zero feature under the CALC menu to determine on what interval(s) 0 eSolutions Manual - Powered by Cognero The zeros of y = 2 cos x – Page 15 are about 0.785 or and about 5.498 or . Therefore, 0 < 2 cos x – on 0 ≤ 5-3 Solving Trigonometric Equations When the area is 36 square inches and the radius is 5 inches, then the measure of θ is 3.01 radians. Solve each inequality on the interval [0, 2 ). 51. 0 < 2 cos x – SOLUTION: on [0, 2π). Use the zero feature under the CALC menu to determine on what interval(s) 0 Graph y = 2 cos x – < 2 cos x – . The zeros of y = 2 cos x – x< or 53. are about 0.785 or and about 5.498 or . Therefore, 0 < 2 cos x – on 0 ≤ < x < 2π. ≤ tan x cot x SOLUTION: Graph y = and y = tan x cot x on [0, 2π). Use the intersect feature under the CALC menu to determine on what interval(s) ≤ tan x cot x. The graphs intersect at about 3.142 or π. Therefore, eSolutions Manual - Powered by Cognero 55. sin x − 1 < 0 SOLUTION: ≤ tan x cot x on 0 ≤ x < 2π. Page 16 The zeros of y = 2 cos x – are about 0.785 or and about 5.498 or . Therefore, 0 < 2 cos x – on 0 ≤ < or Trigonometric < x < 2π. 5-3 xSolving Equations 53. ≤ tan x cot x SOLUTION: and y = tan x cot x on [0, 2π). Use the intersect feature under the CALC menu to determine Graph y = ≤ tan x cot x. on what interval(s) The graphs intersect at about 3.142 or π. Therefore, 55. ≤ tan x cot x on 0 ≤ x < 2π. sin x − 1 < 0 SOLUTION: sin x − 1. Use the zero feature under the CALC menu to determine on what interval(s) Graph y = 1 < 0. The zeros of or sin x − 1 are about 0.785 or and about 2.356 or + n , x = + n , x = + n , and x = = tan x. Vijay thinks that the solutions + n . Alicia thinks that the solutions are x = eSolutions Manual - Powered by Cognero and x = sin x − 1 < 0 on 0 ≤ x < < x < 2π. 57. ERROR ANALYSIS Vijay and Alicia are solving tan2 x – tan x + are x = . Therefore, sin x − + n . Is either of them correct? Explain your reasoning. + n Page 17 sin x − 1 are about 0.785 or The zeros of and about 2.356 or sin x − 1 < 0 on 0 ≤ x < . Therefore, 5-3 Solving Trigonometric Equations < 2π. or < x 57. ERROR ANALYSIS Vijay and Alicia are solving tan2 x – tan x + are x = + n , x = + n , x = + n , and x = and x = + n . Is either of them correct? Explain your reasoning. = tan x. Vijay thinks that the solutions + n . Alicia thinks that the solutions are x = + n SOLUTION: 2 First, solve tan x – tan x + = tan x. and x = On [0, 2π) tan x = 1 when x = and tan x = when x = and x = . Sample answer: Therefore, Vijay’s solutions are correct; however, they are not stated in the simplest form. For + n example, his solutions of x = 1, + nπ is equivalent to and x = + n could simply be stated as x = + n because when n = . CHALLENGE Solve each equation for all values of x. 59. 4 cos2 x – 4 sin2 x cos2 x + 3 sin2 x = 3 SOLUTION: On [0, 2 ) sin x = 1 when x = when x = and x = , sin x = –1 when x = , sin x = – and x = when x = , and sin x = . Therefore, after checking for extraneous solutions, the solutions are eSolutions Manual - Powered by Cognero , , , , , and OPEN ENDED Write a trigonometric equation that has each of the following solutions. . Page 18 Sample answer: Therefore, Vijay’s solutions are correct; however, they are not stated in the simplest form. For + n example, his solutions of x = and x = + n could simply be stated as x = + n because when n = 5-3 1, Solving Trigonometric equivalent to .Equations + nπ is CHALLENGE Solve each equation for all values of x. 59. 4 cos2 x – 4 sin2 x cos2 x + 3 sin2 x = 3 SOLUTION: On [0, 2 ) sin x = 1 when x = when x = and x = , sin x = –1 when x = , sin x = – and x = when x = , and sin x = . Therefore, after checking for extraneous solutions, the solutions are , , , , , and . OPEN ENDED Write a trigonometric equation that has each of the following solutions. 61. SOLUTION: Sample answer: When sin x = 0, x = 0 and x = π. When sin x = solutions of 0, π, , and is 2 = 0 or sin x – ,x= and x = . So, one equation that has 2 = 0. This can be rewritten as 2 sin x = sin x. 63. Writing in Math Explain the difference in the techniques that are used when solving equations and verifying identities. SOLUTION: Sample answer: When solving an equation, you use properties of equality to manipulate each side of the equation to isolate a variable. When verifying an identity, you transform an expression on one side of the identity into the expression on the other side through a series of algebraic steps. eSolutions Manual - Powered by Cognero Verify each identity. 65. = Page 19 solutions of 0, π, , and is 2 = 0 or sin x – 2 = 0. This can be rewritten as 2 sin x = 5-3 Solving sin x. Trigonometric Equations 63. Writing in Math Explain the difference in the techniques that are used when solving equations and verifying identities. SOLUTION: Sample answer: When solving an equation, you use properties of equality to manipulate each side of the equation to isolate a variable. When verifying an identity, you transform an expression on one side of the identity into the expression on the other side through a series of algebraic steps. Verify each identity. 65. = SOLUTION: Find the value of each expression using the given information. 67. tan ; sin θ = , tan > 0 SOLUTION: Use the Pythagorean Identity that involves sin θ. Since tan is positive and sin is positive, cos θ must be positive. So, eSolutions Manual - Powered by Cognero . Page 20 5-3 Solving Trigonometric Equations Find the value of each expression using the given information. 67. tan ; sin θ = , tan > 0 SOLUTION: Use the Pythagorean Identity that involves sin θ. Since tan 69. sec .; tan is positive and sin is positive, cos θ must be positive. So, = –1, sin . < 0 SOLUTION: Use the Pythagorean Identity that involves tan Since tan . is negative and sin θ is negative, cos 2 Given f (x)- Powered = 2x –by5x + 3 and g(x) eSolutions Manual Cognero 71. (f – g)(x) SOLUTION: and = 6x + 4, find each. or sec θ must be positive. Therefore, . Page 21 5-3 Since Solving Equations tan Trigonometric negative, cos is negative and sin θ is and or sec θ must be positive. Therefore, . 2 Given f (x) = 2x – 5x + 3 and g(x) = 6x + 4, find each. 71. (f – g)(x) SOLUTION: 73. (x) SOLUTION: 75. SAT/ACT For all positive values of m and n, if = 2, then x = A B C D E SOLUTION: The correct answer is D. 77. Which of the following is not eSolutions Manual - Powered by Cognero A a solution of 0 = sin 2 + cos tan ? Page 22 5-3 Solving Trigonometric Equations The correct answer is D. 77. Which of the following is not a solution of 0 = sin 2 + cos tan ? A B C 2π D SOLUTION: Try choice A. Try choice B. Try choice C. Try choice D. The correct answer is D. eSolutions Manual - Powered by Cognero Page 23