5-3 Solving Trigonometric Equations

Transcription

1 Solve each equation for all values of x sin x + 2 = sin x The period of sine is 2π, so you only need to find solutions on the interval. The solutions on this interval are and. Solutions on the interval (, ), are found by adding integer multiples of 2π. Therefore, the general form of the solutions is + 2nπ, + 2nπ, = 4 cos 2 x + 1 The period of cosine is 2π, so you only need to find solutions on the interval. The solutions on this interval are,,, and. Solutions on the interval (, ), are found by adding integer multiples of 2π. Therefore, the general form of the solutions is + 2nπ, + 2nπ, + 2nπ, + 2nπ,. esolutions Manual - Powered by Cognero Page 1

2 cot 2 x = 12 The period of cotangent is π, so you only need to find solutions on the interval. The solutions on this interval are and. Solutions on the interval (, ), are found by adding integer multiples of π. Therefore, the general form of the solutions is + nπ, + nπ, csc x = 2 csc x + The period of cosecant is 2π, so you only need to find solutions on the interval. The solutions on this interval are and. Solutions on the interval (, ), are found by adding integer multiples of 2π. Therefore, the general form of the solutions is + 2nπ, + 2nπ, tan 2 x 2 = 4 The period of tangent is π, so you only need to find solutions on the interval. The solutions on this interval are and. Solutions on the interval (, ), are found by adding integer multiples of π. Therefore, the general form of the solutions is + nπ, + nπ,. esolutions Manual - Powered by Cognero Page 2

3 11. 7 cot x = 4 cot x The period of cotangent is π, so you only need to find solutions on the interval. The only solution on this interval is. Solutions on the interval (, ), are found by adding integer multiples of π. Therefore, the general form of the solutions is + nπ,. Find all solutions of each equation on [0, 2 ). 13. sin 4 x + 2 sin 2 x 3 = 0 On the interval [0, 2π), when x = and when x =. Since is not a real number, the equation yields no additional solutions cot x = cot x sin 2 x 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. esolutions Manual - Powered by Cognero Page 3

4 17. cos 3 x + cos 2 x cos x = 1 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. esolutions Manual - Powered by Cognero Page 4

5 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 = v 0 2 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. 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 esolutions Manual - Powered by Cognero Page 5

6 Find all solutions of each equation on the interval [0, 2 ) = cot 2 x + csc x Therefore, on the interval [0, 2π) the solutions are,, and. 23. tan 2 x = 1 sec x Therefore, on the interval [0, 2π) the solutions are 0,, and. esolutions Manual - Powered by Cognero Page 6

7 cos 2 x = sin x + 1 Therefore, on the interval [0, 2π) the solutions are,, and sin x = 3 3 cos x Therefore, on the interval [0, 2π) the only valid solutions are and 0. esolutions Manual - Powered by Cognero Page 7

8 29. sec 2 x 1 + tan x tan x = Therefore, on the interval [0, 2π) the solutions are,, and. esolutions Manual - Powered by Cognero Page 8

9 31. OPTOMETRY Optometrists sometimes join two oblique or tilted prisms to correct vision. The resultant refractive power P R of joining two oblique prisms can be calculated by first resolving each prism into its horizontal and vertical components, P H and P V. Using the equations above, determine for what values of P V and P H are equivalent. The sine and cosine have the same values in the interval [0, 2π) at and. Therefore, the components will be equivalent when. esolutions Manual - Powered by Cognero Page 9

10 Find all solutions of each equation on the interval [0, 2 ) = 4 On the interval [0, 2π),cos x = when x = and when x =. esolutions Manual - Powered by Cognero Page 10

11 35. cot x cos x + 1 = + On, when x = and when x =. GRAPHING CALCULATOR Solve each equation on the interval [0, 2 nearest hundredth. 37. sin x + cos x = 3x ) by graphing. Round to the On the interval, the only solution is when x = esolutions Manual - Powered by Cognero Page 11

12 39. x log x + 5x cos x = 2 On the interval, the solutions are when x = 1.84 and when x = Find the x-intercepts of each graph on the interval [0, 2 ). 41. Let y = 0 and solve for x. On the interval [0, 2π) cos x = 0 when x = and x =. esolutions Manual - Powered by Cognero Page 12

13 43. Let y = 0 and solve for x. On the interval [0, 2π) cot x = 1 when x = and x =. Find all solutions of each equation on the interval [0, 4 ) tan x = 2 sec 2 x On the interval [0, 4π) the solutions are,,, and. esolutions Manual - Powered by Cognero Page 13

14 47. csc x cot 2 x = csc x 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. 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 Page 14

15 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. esolutions Manual - Powered by Cognero Page 15

16 Solve each inequality on the interval [0, 2 ) < 2 cos x Graph y = 2 cos x on [0, 2π). Use the zero feature under the CALC menu to determine on what interval(s) 0 < 2 cos x. The zeros of y = 2 cos x are about or and about or. Therefore, 0 < 2 cos x on 0 x < or < x < 2π. esolutions Manual - Powered by Cognero Page 16

17 53. tan x cot x 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 or π. Therefore, tan x cot x on 0 x < 2π. 55. sin x 1 < 0 Graph y = sin x 1. Use the zero feature under the CALC menu to determine on what interval(s) sin x 1 < 0. The zeros of sin x 1 are about or and about or. Therefore, sin x 1 < 0 on 0 x < or < x < 2π. esolutions Manual - Powered by Cognero Page 17

18 57. ERROR ANALYSIS Vijay and Alicia are solving tan 2 x tan x + = tan x. Vijay thinks that the solutions are x = + n, x = + n, x = + n, and x = + n. Alicia thinks that the solutions are x = + n and x = + n. Is either of them correct? Explain your reasoning. First, solve tan 2 x tan x + = tan x. On [0, 2π) tan x = 1 when x = and 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 example, his solutions of x = + n and x = + n could simply be stated as x = + n because when n = 1, + nπ is equivalent to. esolutions Manual - Powered by Cognero Page 18

19 CHALLENGE Solve each equation for all values of x cos 2 x 4 sin 2 x cos 2 x + 3 sin 2 x = 3 On [0, 2 ) sin x = 1 when x =, sin x = 1 when x =, sin x = when x = and x =, and sin x = when x = and 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. Sample answer: When sin x = 0, x = 0 and x = π. When sin x =, x = and x =. So, one equation that has solutions of 0, π,, and is = 0 or sin 2 x = 0. This can be rewritten as 2 sin 2 x = sin x. esolutions Manual - Powered by Cognero Page 19

20 63. Writing in Math Explain the difference in the techniques that are used when solving equations and verifying identities. 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. = esolutions Manual - Powered by Cognero Page 20

21 Find the value of each expression using the given information. 67. tan ; sin θ =, tan > 0 Use the Pythagorean Identity that involves sin θ. Since tan is positive and sin is positive, cos θ must be positive. So,. 69. sec.; tan = 1, sin < 0 Use the Pythagorean Identity that involves tan. Since tan is negative and sin θ is negative, cos and or sec θ must be positive. Therefore,. esolutions Manual - Powered by Cognero Page 21

22 Given f (x) = 2x 2 5x + 3 and g(x) = 6x + 4, find each. 71. (f g)(x) 73. (x) 75. SAT/ACT For all positive values of m and n, if = 2, then x = A B C D E The correct answer is D. esolutions Manual - Powered by Cognero Page 22

23 77. Which of the following is not a solution of 0 = sin + cos tan 2? A B C 2π D Try choice A. Try choice B. Try choice C. Try choice D. The correct answer is D. esolutions Manual - Powered by Cognero Page 23