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Trigonometric Equations. In quadratic form, using identities or linear in sine and cosine. Solving a Trig Equation in Quadratic Form. Solve the equation: 2sin 2 θ – 3 sin θ + 1 = 0, 0 ≤ θ ≤ 2 p Let sin θ equal some variable sin θ = a Factor this equation (2a – 1) (a – 1) = 0

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trigonometric equations

Trigonometric Equations

In quadratic form, using identities or linear in sine and cosine

solving a trig equation in quadratic form
Solving a Trig Equation in Quadratic Form
  • Solve the equation:
  • 2sin2θ – 3 sin θ + 1 = 0, 0 ≤ θ≤ 2p
  • Let sin θ equal some variable
  • sin θ = a
  • Factor this equation
  • (2a – 1) (a – 1) = 0
  • Therefore a = ½ a = 1
solving a trig equation in quadratic form1
Solving a Trig Equation in Quadratic Form
  • Now substitute sin θ back in for a
  • sin θ = ½ sin θ = 1
  • Now do the inverse sin to find what θ equals
  • θ = sin-1 (½) θ = sin-1 1
  • θ = p/6 and 5p/6 θ = p/2
solving a trig equation in quadratic form2
Solving a Trig Equation in Quadratic Form
  • Solve the equation:
  • (tan θ – 1)(sec θ – 1) = 0
  • tan θ – 1 = 0 sec θ – 1 = 0
  • tan θ = 1 sec θ = 1
  • θ = tan-1 1 θ = sec-1 1
  • θ = p/4 and 5p/4 θ = 0
solving a trig equation using identities
Solving a Trig Equation Using Identities
  • In order to solve trig equations, we want to have a single trig word in the equation. We can use trig identities to accomplish this goal.
  • Solve the equation
  • 3 cos θ + 3 = 2 sin2θ
  • Use the pythagorean identities to change sin2 θ to cos θ
solving a trig equation using identities1
Solving a Trig Equation Using Identities
  • sin2 = 1 – cos2 θ
  • Substituting into the equation
  • 3 cos θ + 3 = 2(1 – cos2θ)
  • To solve a quadratic equation it must be equal to 0
  • 2cos2θ + 3 cos θ + 1 = 0
  • Let cos θ = b
solving a trig equation using identities2
Solving a Trig Equation using Identities
  • 2b2 + 3b + 1 = 0
  • (2b + 1) (b + 1) = 0
  • (2b + 1) = 0 b + 1 = 0
  • b = -½ b = -1
  • cos θ = -½ cos θ = -1
  • θ = 2p/3, 4p/3 θ = p
solving a trig equation using identities3
Solving a Trig Equation Using Identities
  • cos2θ – sin2θ + sin θ = 0
  • 1 – sin2θ – sin2θ + sin θ = 0
  • -2sin2 θ + sin θ + 1 = 0
  • 2 sin2θ – sin θ – 1 = 0
  • Let c = sin θ
  • 2c2 – c – 1 = 0
  • (2c + 1) (c – 1) = 0
solving a trig equation using identities4
Solving a Trig Equation Using Identities
  • (2c + 1) = 0 c – 1 = 0
  • c = -½ c = 1
  • sin θ = -½ sin θ = 1
  • θ = p/3 + p q=2p-p/3θ = p/2
  • θ = 4p/3, q = 7p/3
solving a trig equation using identities5
Solving a Trig Equation Using Identities
  • Solve the equation
  • sin (2θ) sin θ = cos θ
  • Substitute in the formula for sin 2θ
  • (2sin θ cos θ)sin θ=cos θ
  • 2sin2 θ cos θ – cos θ = 0
  • cos θ(2sin2 – 1) = 0
  • cos θ = 0 2sin2 θ=1
solving a trig equation using identities6
Solving a Trig Equation Using Identities
  • cos θ = 0
  • θ = 0, pθ = p/4, 3p/4, 5p/4, 7p/4
solving a trig equation using identities7
Solving a Trig Equation Using Identities
  • sin θ cos θ = -½
  • This looks very much like the sin double angle formula. The only thing missing is the two in front of it.
  • So . . . multiply both sides by 2
  • 2 sin θ cos θ = -1
  • sin 2θ = -1
  • 2 θ = sin-1 -1
solving a trig equation using identities8
Solving a Trig Equation Using Identities
  • 2θ = 3p/2
  • θ = 3p/4 2θ = 3p/2 + 2p

2q = 7p/2

q = 7p/4

solving a trig equation linear in sin and cos
Solving a Trig Equation Linear in sin θ and cos θ
  • sin θ + cos θ = 1
  • There is nothing I can substitute in for in this problem. The best way to solve this equation is to force a pythagorean identity by squaring both sides.
  • (sin θ + cos θ)2 = 12
solving a trig equation linear in sin and cos1
Solving a Trig Equation Linear in sin θ and cos θ
  • sin2θ + 2sin θ cos θ + cos2 θ = 1
  • 2sin θ cos θ + 1 = 1
  • 2sin θ cos θ = 0
  • sin 2θ = 0
  • 2θ = 0 2θ = p
  • θ = 0 θ = p/2
  • θ = p θ = 3p/2
solving a trig equation linear in sin and cos2
Solving a Trig Equation Linear in sin θ and cos θ
  • Since we squared both sides, these answers may not all be correct (when you square a negative number it becomes positive).
  • In the original equation, there were no terms that were squared
solving a trig equation linear in sin and cos3
Solving a Trig Equation Linear in sin θ and cos θ
  • Check:
  • Does sin 0 + cos 0 = 1?
  • Does sin p/2 + cos p/2 = 1?
  • Does sin p + cos p = 1?
  • Does sin 3p/2 + cos 3p/2 = 1?
solving a trig equation linear in sin and cos4
Solving a Trig Equation Linear in sin θ and cos θ
  • sec θ = tan θ + cot θ
  • sec2 θ = (tan θ + cot θ)2
  • sec2 θ = tan2 θ + 2 tan θ cot θ + cot2 θ
  • sec2 θ = tan2 θ + 2 + cot2 θ
  • sec2 θ – tan2 θ = 2 + cot2 θ
  • 1 = 2 + cot2 θ
  • -1 = cot2 θ
solving a trig equation linear in sin and cos5
Solving a Trig Equation Linear in sin θ and cos θ
  • q is undefined (can’t take the square root of a negative number).
solving trig equations using a graphing utility
Solving Trig Equations Using a Graphing Utility
  • Solve 5 sin x + x = 3. Express the solution(s) rounded to two decimal places.
  • Put 5 sin x + x on y1
  • Put 3 on y2
  • Graph using the window 0 ≤ θ ≤2p
  • Find the intersection point(s)
word problems
Word Problems
  • Page 519 problem 58