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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: 2sin2 θ – 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

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**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

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**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 θ = sec-1 1 θ = p/4 and 5p/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 sin2 θ Use the pythagorean identities to change sin2 θ to cos θ

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**Solving a Trig Equation Using Identities**

sin2 = 1 – cos2 θ Substituting into the equation 3 cos θ + 3 = 2(1 – cos2 θ) To solve a quadratic equation it must be equal to 0 2cos2 θ + 3 cos θ + 1 = 0 Let cos θ = b

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**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

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**Solving a Trig Equation Using Identities**

cos2 θ – sin2 θ + sin θ = 0 1 – sin2 θ – sin2 θ + sin θ = 0 -2sin2 θ + sin θ + 1 = 0 2 sin2 θ – sin θ – 1 = 0 Let c = sin θ 2c2 – c – 1 = 0 (2c + 1) (c – 1) = 0

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**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/3 θ = p/2 θ = 4p/3, q = 7p/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 2sin2 θ cos θ – cos θ = 0 cos θ(2sin2 – 1) = 0 cos θ = 0 2sin2 θ=1

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**Solving a Trig Equation Using Identities**

cos θ = 0 θ = 0, p θ = p/4, 3p/4, 5p/4, 7p/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 = 3p/2 θ = 3p/4 2θ = 3p/2 + 2p 2q = 7p/2 q = 7p/4

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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 = 12

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**Solving a Trig Equation Linear in sin θ and cos θ**

sin2 θ + 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

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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 p/2 + cos p/2 = 1? Does sin p + cos p = 1? Does sin 3p/2 + cos 3p/2 = 1?

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**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 cot2 θ sec2 θ – tan2 θ = 2 + cot2 θ 1 = 2 + cot2 θ -1 = cot2 θ

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**Solving a Trig Equation Linear in sin θ and cos θ**

q is undefined (can’t 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 y1 Put 3 on y2 Graph using the window 0 ≤ θ ≤2p Find the intersection point(s)

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Word Problems Page 519 problem 58

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Trigonometric Equations M 140 Precalculus V. J. Motto.

Trigonometric Equations M 140 Precalculus V. J. Motto.

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