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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 racquet 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 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 racquet 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 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 ). 2 21. 1 = Manual cot x- +Powered csc x by Cognero eSolutions 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 eSolutions Manual - Powered by Cognero , , and . Page 6