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

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Solving a Trig Equation in Quadratic Form Solve the equation: 2sin 2 θ – 3 sin θ + 1 = 0, 0 θ 2 Let sin θ equal some variable sin θ = a Factor this equation (2a – 1) (a – 1) = 0 Therefore a = ½ a = 1

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Solving a Trig Equation in Quadratic Form Now substitute sin θ back in for a sin θ = ½sin θ = 1 Now do the inverse sin to find what θ equals θ = sin -1 (½) θ = sin -1 1 θ = /6 and 5/6 θ = /2

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Solving a Trig Equation in Quadratic Form Solve the equation: (tan θ – 1)(sec θ – 1) = 0 tan θ – 1 = 0sec θ – 1 = 0 tan θ = 1sec θ = 1 θ = tan -1 1 θ = sec -1 1 θ = /4 and 5/4 θ = 0

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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 sin 2 θ Use the pythagorean identities to change sin 2 θ to cos θ

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Solving a Trig Equation Using Identities sin 2 = 1 – cos 2 θ Substituting into the equation 3 cos θ + 3 = 2(1 – cos 2 θ) To solve a quadratic equation it must be equal to 0 2cos 2 θ + 3 cos θ + 1 = 0 Let cos θ = b

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Solving a Trig Equation using Identities 2b 2 + 3b + 1 = 0 (2b + 1) (b + 1) = 0 (2b + 1) = 0b + 1 = 0 b = -½b = -1 cos θ = -½cos θ = -1 θ = 2/3, 4/3 θ =

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Solving a Trig Equation Using Identities cos 2 θ – sin 2 θ + sin θ = 0 1 – sin 2 θ – sin 2 θ + sin θ = 0 -2sin 2 θ + sin θ + 1 = 0 2 sin 2 θ – sin θ – 1 = 0 Let c = sin θ 2c 2 – c – 1 = 0 (2c + 1) (c – 1) = 0

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Solving a Trig Equation Using Identities (2c + 1) = 0c – 1 = 0 c = -½c = 1 sin θ = -½sin θ = 1 θ = + θ = θ = 4/3,

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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 2sin 2 θ cos θ – cos θ = 0 cos θ(2sin 2 – 1) = 0 cos θ = 02sin 2 θ=1

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Solving a Trig Equation Using Identities cos θ = 0 θ = 0, θ = /4,

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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

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Solving a Trig Equation Using Identities 2θ = 3 θ = 3/4 2θ = 3

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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 = 1 2

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Solving a Trig Equation Linear in sin θ and cos θ sin 2 θ + 2sin θ cos θ + cos 2 θ = 1 2sin θ cos θ + 1 = 1 2sin θ cos θ = 0 sin 2θ = 0 2θ = 2θ = θ = θ = θ = θ = 3

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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

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Solving a Trig Equation Linear in sin θ and cos θ Check: Does sin 0 + cos 0 = 1? Does sin cos Does sin 3cos 3

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Solving a Trig Equation Linear in sin θ and cos θ sec θ = tan θ + cot θ sec 2 θ = (tan θ + cot θ) 2 sec 2 θ = tan 2 θ + 2 tan θ cot θ + cot 2 θ sec 2 θ = tan 2 θ cot 2 θ sec 2 θ – tan 2 θ = 2 + cot 2 θ 1 = 2 + cot 2 θ -1 = cot 2 θ

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Solving a Trig Equation Linear in sin θ and cos θ is undefined (cant take the square root of a negative number).

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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 y 1 Put 3 on y 2 Graph using the window 0 θ 2 Find the intersection point(s)

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Word Problems Page 519 problem 58

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