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14-6Solving 14-6 SolvingTrigonometric TrigonometricEquations Equations Warm Up Lesson Presentation Lesson Quiz Holt Algebra Holt Algebra 22 14-6 Solving Trigonometric Equations Warm Up Solve. 1. x2 + 3x – 4 = 0 x = 1 or – 4 2. 3x2 + 7x = 6 Evaluate each inverse trigonometric function. 3. Tan-1 1 45° 4. Sin-1 – 60 Holt Algebra 2 14-6 Solving Trigonometric Equations Objective Solve equations involving trigonometric functions. Holt Algebra 2 14-6 Solving Trigonometric Equations Unlike trigonometric identities, most trigonometric equations are true only for certain values of the variable, called solutions. To solve trigonometric equations, apply the same methods used for solving algebraic equations. Holt Algebra 2 14-6 Solving Trigonometric Equations Example 1: Solving Trigonometric Equations with Infinitely Many Solutions Find all the solutions of sinθ = sinθ + Method 1 Use algebra. Solve for θ over the principal value of sine, –90° ≤ θ ≤ 90°. sinθ = sinθ + sinθ sinθ = Subtract sinθ = Combine like terms. Holt Algebra 2 sinθ from both sides. 14-6 Solving Trigonometric Equations Example 1 Continued sinθ = Multiply by 2. θ = sin-1 Apply the inverse sineθ. θ = 30° Find θ when sinθ = Find all real number value of θ, where n is an integer. θ = 30° + 360°n θ = 150° + 360°n Holt Algebra 2 Use the period of the sine function. Use reference angles to find other values of θ. 14-6 Solving Trigonometric Equations Example 1 Continued Method 2 Use a graph. 1 Graph y = sinθ and y = sinθ + in the same viewing window for –90° ≤ θ ≤ 90°. 90 –90 Use the intersect feature of your graphing calculator to find the points of intersection. –1 The graphs intersect at θ = 30°. Thus, θ = 30° + 1360°n, where n is an integer. Holt Algebra 2 14-6 Solving Trigonometric Equations Check It Out! Example 1 Find all of the solutions of 2cosθ + = 0. Method 1 Use algebra. Solve for θ over the principal value of sine, 0 ≤ θ ≤ . 2cosθ = cosθ = Holt Algebra 2 Subtract from both sides. Divide both sides by 2. θ = cos-1 – Apply the inverse cosineθ. θ = 150° Find θ when cosine θ = . 14-6 Solving Trigonometric Equations Check It Out! Example 1 Continued θ = 150° + 360°n, 210° +360°n. Use reference angles to find other values of θ. Method 2 Use a graph. 2 Graph y = 2cosθ and y= in the same –360 360 viewing window for –360° ≤ θ ≤ 360°. –2 The graphs intersect at θ = 150°. Thus, θ = 150° + 360°n, where n is an integer. Holt Algebra 2 14-6 Solving Trigonometric Equations Some trigonometric equations can be solved by applying the same methods used for quadratic equations. Holt Algebra 2 14-6 Solving Trigonometric Equations Example 2A: Solving Trigonometric Equations in Quadratic Form Solve each equation for the given domain. 4tan2θ – 7 tanθ + 3 = 0 for 0° ≤ θ ≤ 360°. 4tan2θ – 7 tanθ + 3 = 0 (tanθ – 1)(4tanθ – 3) = 0 Holt Algebra 2 Factor the quadratic expression by comparing it with 4x2 – 7x + 3 = 0. Apply the Zero Product Property. 14-6 Solving Trigonometric Equations Example 2A Continued tanθ = 1 or tan θ = θ = tan-1(1) = 45° or 225° Holt Algebra 2 θ = tan-1 Apply the inverse tangent. Use a calculator. Find all angles ≈ 36.9° or 216.9° for 0°≤ θ ≤360°. 14-6 Solving Trigonometric Equations Example 2B: Solving Trigonometric Equations in Quadratic Form 2cos2θ – cosθ = 1 for 0 ≤ θ ≤ . 2cos2θ – cosθ – 1 = 0 (2cosθ + 1) (cosθ – 1) = 0 cosθ = θ= Holt Algebra 2 or cosθ = 1 or θ = 0 Subtract 1 from both sides. Factor the quadratic expression by comparing it with 2x2 – x + 1 = 0. Apply the Zero Product Property. Find both angles for 0 ≤ θ ≤ . 14-6 Solving Trigonometric Equations Check It Out! Example 2a Solve each equation for 0 ≤ θ ≤ 2. cos2 θ + 2cosθ = 3 cos2 θ + 2cosθ – 3 = 0 (cosθ – 1)(cosθ + 3) = 0 Subtract 3 from both sides. Factor the quadratic expression by comparing it to x2 +2x – 3 = 0. cosθ = 1 or cosθ = –3 Apply the Zero Product Property. cosθ = – 3 has no solution because –3 ≤ cosθ ≤ 1. cosθ = 2 or 0 Holt Algebra 2 The only solution will come from cosθ = 1. 14-6 Solving Trigonometric Equations Check It Out! Example 2b Solve each equation for 0 ≤ θ ≤ 2. sin2θ + 5 sinθ – 2 = 0 The equation is in quadratic form but can not be easily factored. Use the quadratic formula. sinθ = Holt Algebra 2 14-6 Solving Trigonometric Equations Check It Out! Example 2b Continued Apply the inverse sine. Use a calculator. Find both angles. Holt Algebra 2 14-6 Solving Trigonometric Equations You can often write trigonometric equations involving more than one function as equations of only one function by using trigonometric identities. Holt Algebra 2 14-6 Solving Trigonometric Equations Example 3A: Solving Trigonometric Equations with Trigonometric Identities Use trigonometric identities to solve each equation. tan2θ + sec2θ = 3 for 0 ≤ θ ≤ 2π. 2 2 tan2θ + (1 + tan2θ) – 3 = 0 Substitute 1 + tan θ for sec θ by the Pythagorean identity. 2tan2θ – 2 = 0 tan2θ – 1 = 0 Simplify. Divide by 2. (tanθ – 1)(tanθ + 1) = 0 Factor. Apply the Zero Product tanθ = 1 or tanθ = – 1 Property. Holt Algebra 2 14-6 Solving Trigonometric Equations Example 3A Continued Check Use the intersect feature of your graphing calculator. A graph supports your answer. Holt Algebra 2 14-6 Solving Trigonometric Equations Example 3B: Solving Trigonometric Equations with Trigonometric Identities Use trigonometric identities to solve each equation. cos2θ = 1 + sin2θ for 0° ≤ θ ≤ 360° (1 – sin2θ) – 1– sin2θ = 0 Substitute 1 – sin2θ for cos2θ by the Pythagorean identity. –2sin2θ = 0 Simplify. sin2θ = 0 Divide both sides by – 2. sinθ = 0 Take the square root of both sides. θ = 0° or 180° or 360° Holt Algebra 2 14-6 Solving Trigonometric Equations Example 3B Continued cos2θ = 1+sin2θ for 0° ≤ θ ≤ 360° θ = 0° or 180° or 360° Check Use the intersect feature of your graphing calculator. A graph supports your answer. Holt Algebra 2 14-6 Solving Trigonometric Equations Check It Out! Example 3a Use trigonometric identities to solve each equation for the given domain. 4sin2θ + 4cosθ = 5 2θ for sin2θ Substitute 1 – cos 4(1 - cos2θ) + 4cosθ – 5 = 0 by the Pythagorean identity. 4cos2θ – 4cosθ + 1 = 0 Simplify. (2cos2θ – 1)2 = 0 Factor. Take the square root of both sides and simplify. Holt Algebra 2 14-6 Solving Trigonometric Equations Check It Out! Example 3b Use trigonometric identities to solve each equation for the given domain. sin2θ = – cosθ for 0 ≤ θ < 2 2cosθsinθ + cosθ = 0 cosθ(2sinθ + 1) = 0 Substitute 2cosθsinθ for sin2θ by the double-angle identity. Factor. Apply the Zero Product Property. Holt Algebra 2 14-6 Solving Trigonometric Equations Example 4: Problem-Solving Application On what days does the sun rise at 4 A.M. on Cadillac Mountain? The time of the sunrise can be modeled by Holt Algebra 2 14-6 Solving Trigonometric Equations 1 Understand the Problem The answer will be specific dates in the year. List the important information: • The function model is t(m) = 1.665 (m + 3) + 5.485. • Sunrise is at 4 A.M., which is represented by t = 4. • m represents the number of months after January 1. Holt Algebra 2 14-6 Solving Trigonometric Equations 2 Make a Plan Substitute 4 for t in the model. Then solve the equation for m by using algebra. 3 Solve 4 = 1.665sin (m + 3) + 5.485 Substitute 4 for t. Isolate the sine term. sin-1(–0.8918) = Holt Algebra 2 (m + 3) Apply the inverse sine θ. 14-6 Solving Trigonometric Equations Sine is negative in Quadrants lll and lV. Compute both values. Qlll: π + sin-1(0.8918) Holt Algebra 2 QlV: 2π + sin-1(0.8918) 14-6 Solving Trigonometric Equations Using an average of 30 days per month, the date m = 5.10 corresponds to June 4(5 months and 3 days after January 1) and m = 6.90 corresponds to July 28 (6 months and 27 days after January 1). Holt Algebra 2 14-6 Solving Trigonometric Equations 4 Look Back Check your answer by using a graphing calculator. Enter y = 1.665sin (x + 3) + 5.485 and y = 4. Graph the functions on the same viewing window, and find the points of intersection. The graphs intersect at early June and late July. Holt Algebra 2 14-6 Solving Trigonometric Equations Check It Out! Example 4 The number of hours h of sunlight in a day at Cadillac Mountain can be modeled by h(d) = 3.31sin (d – 85.25) + 12.22, where d is the number of days after January 1. When are there 12 hours of sunlight. 1 Understand the Problem The answer will be specific dates in the year. Holt Algebra 2 14-6 Solving Trigonometric Equations 1 Understand the Problem The answer will be specific dates in the year. List the important information: • The function model is h(d) = 3.31sin (d – 85.25) + 12.22. • The number of hours of sunlight in the day, which is represented by h = 12. • d represents the number of days after January 1. Holt Algebra 2 14-6 Solving Trigonometric Equations 2 Make a Plan Substitute 12 for h in the model. Then solve the equation for d by using algebra. 3 Solve 12 = 3.31sin (d – 85.25) + 12.22 Substitute 12 for h. Isolate the sine term. Apply the inverse sine θ. Holt Algebra 2 14-6 Solving Trigonometric Equations Sine is negative in Quadrants lll and lV. Compute both values. Qlll: 81.4 ≈ d Holt Algebra 2 14-6 Solving Trigonometric Equations QlV: 271.6 ≈ d Holt Algebra 2 14-6 Solving Trigonometric Equations 4 Look Back Check your answer by using a graphing calculator. Enter y = 3.31sin (d – 85.25) + 12.22 Graph the functions on the same viewing window, and find the points of intersection. The graphs intersect in late March and late September. Holt Algebra 2 14-6 Solving Trigonometric Equations Lesson Quiz 1. Find all solutions for cosθ = – cosθ. θ = 45° + n 360° or 315° + n 360° 2. Solve 3sin2θ – 4 = 0 for 0 ≤ θ ≤ 360°. θ ≈ 221.8° or 318.2° 3. Solve cos2θ = 3sinθ + 2 for 0 ≤ 0 ≤ 2π. Holt Algebra 2