Solving Trig Equations

Παρουσίαση με θέμα: "Solving Trig Equations"— Μεταγράφημα παρουσίασης:

1 Solving Trig Equations

2 Solve. 2 sin θ = 1 sin θ = ½ θ = π/6 OR θ = 5 π/6 But the period of sin is 2π, so θ = π/6 + 2πk θ = 5 π/6 + 2πk Think of how these multiple solutions would appear on a graph.

3 Solve. sec x = 2.85 so cos x = .351 Using a calculator (in radian mode), x = Also, 2π – = is a solution. So x = πk or x = πk

4 Solve. cos2x + cos x – 2 = 0 (cos x + 2)(cos x – 1) = 0 cos x = -2 or cos x = 1 no solution x = 0 + 2πk = 2πk

5 Solve. 3tan2x – sec2x – 5 = 0 3tan2x - (tan2x + 1) – 5 = 0 2tan2x - 6 = 0 tan2x = 3 tan x = ±√3 x = π/3 or x = 2π/3 , but the period is π, so x= π/3 + πk or x = 2π/3 + πk (or could give all 4 answers)

6 Solve. sin θ = 2cos θ tan θ = 2 θ = 1.107 but the period is π, so θ = πk

7 Solve. sinx – 2 = 3csc x sin x – 2 = 3/sin x sin2 x – 2sin x = 3 sin2 x – 2sin x – 3 = 0 (sin x – 3)(sinx + 1) = 0 x = 3π/2 + 2πk

8 Solve. sin x + cos x = 1 (sin x + cos x)2 = 12 sin2x + 2 sin x cos x + cos2x = 1 2 sinx cos x = 0 sin x = 0 or cos x = 0 x = 0 or π or x = π/2 or 3π/2 Which of these work? So x =0 +2πk = 2πk or x = π/2 + 2πk

9 Why can’t you divide both sides by 3? Since sin π/2 = 1 ,
Solve . sin 3x = 1 Why can’t you divide both sides by 3? Since sin π/2 = 1 , 3x = π/2 and therefore x = π/6 is a solution. But all solutions are in the form π/2 + 2πk, so 3x = π/2 + 2πk and x = π/6 + (2πk)/3 Can you list all solutions between 0 and 2π?