Aim: Solving Quadratic Trig Equations Course: Alg. 2 & Trig. Aim: How do we solve quadratic trigonometric equations? Do Now: Solve by factoring: x 2 –

Aim: Solving Quadratic Trig Equations Course: Alg. 2 & Trig.

Aim: How do we solve quadratic trigonometric equations?

Do Now:

Solve by factoring:x2 – 3x – 4 = 0


Solving Quadratic Equations

Solve by factoring: x2 – 3x – 4 = 0

x2 – 3x – 4 = 0(x – 4)(x + 1) = 0

(x – 4) = 0 (x + 1) = 0

x = 4 x = -1

Solve using quadratic formula: x2 – 3x – 4 = 0

x b b2 4ac

2a a = 1, b = -3, c = -4

x ( 3) ( 3)2 4(1)( 4)

2(1)

x 3 9 16

2

x 3 25

2x = 4 or -1


Quadratic Trig Equations - Factoring

tan2 – 3tan – 4 = 0

(tan – 4)(tan + 1) = 0

(tan – 4) = 0 (tan + 1) = 0

tan = 4 tan = -1

Solve for by factoring:to nearest degree, in the interval 0º ≤ ≤ 360º

tan is (–) in QII & QIV; reference = 45º

tan is (+) in QI & QIII; reference = 76º

QII 180 – 45 = 135º

QIV 360 – 45 = 315º

and

QI 76º

QIII 180 + 76 = 256º

and

{76º, 135º, 256º, 315º}

= 45º & 76º


Quadratic Trig Equations - Formula

tan2 – 3tan – 4 = 0to nearest degree, in the interval 0º ≤ ≤ 360º



QII 180 – 45 = 135º

QIV 360 – 45 = 315º

andQI 76º

QIII 180 + 76 = 256º

and

{76º, 135º, 256º, 315º}

tan b b2 4ac

2aa = 1, b = -3, c = -4

tan ( 3) ( 3)2 4(1)( 4)

2(1)

tan 3 9 16

2

tan 3 25

2tan = 4 and -1 = 45º & 76º


Model Problem

cos b b 2 4ac

2aa = 3, b = -5, c = -4

cos ( 5) ( 5)2 4(3)( 4)

2(3)

cos 5 73

62.25733

Given: 3cos2 – 5cos – 4 = 0, find to thenearest degree in the interval 0º ≤ ≤ 360º

cos 5 25 48

6

cos 5 73

6 .590667

Calculator

2nd 73 ENTER

Display: -.59066722909

( 5 –

) ) ÷ 6

2nd ENTER

Display: 53.7956245

(–) ANSCOS-1 2nd

= 54º


Model Problem (Con’t)

cos b b 2 4ac

2aa = 3, b = -5, c = -4

Given: 3cos2 – 5cos – 4 = 0, find to thenearest degree in the interval 0º ≤ ≤ 360º

= 54º

cosine is (–) in QII & QIII; reference = 54º

QII

180 + 54 = 234º

and

QIII

cos 5 73

6 .590667

180 – 54 = 126º

{126º, 234º}


Special Quadratics

Solve for in the interval 0º ≤ ≤ 360º: tan2 – 3 = 0

= 60º

tan2 = 3

tan 3



QII 180 – 60 = 120º

QIV 360 – 60 = 300º

andQI 60º

QIII 180 + 60 = 240º

and

{60º, 120º, 240º, 300º}


Model Problem

Solve the equation 2cos2 = cos , for all values of in the interval 0º ≤ ≤ 360º

standard form: 2cos2 – cos = 0cos(2cos – 1) = 0

cos = 0 (2cos – 1) = 0

factor & solve:

2cos = 1cos = 1/2 or .5

= 90º and 270º = 60º

cosine is (+) in QI & QIV; reference = 60º

QI 60º QIV 300 – 60 = 300ºand

{60º, 90º, 270º, 300º}


Model Problem

a = 2, b = -1, c = 02( 1) ( 1) 4(1)(0)

2(2)x

1 1 0

4x

2 0,

4 4x x = .5 and 0

Solve the equation 2cos2 = cos , for all values of in the interval 0º ≤ ≤ 360º

2cos2 – cos = 0

= 90º and 270º = 60º

cosine is (+) in QI & QIV; reference = 60º

QI 60º QIV 300 – 60 = 300ºand

{60º, 90º, 270º, 300º}


Regents Prep

Find all values of in the interval 0 < < 360o that satisfy the equation

2 sin2 + sin = 1.


Model Problem

a = 3, b = 4, c = -1

In the interval 0º ≤ ≤ 360º, find to the nearestdegree, for all values of that satisfy the equation

3sin 4 1

sinrewrite w/o fractions:

3sin2 4sin 1

standard form:

3sin2 4sin 1 0

sin b b2 4ac

2a

sin 4 42 4(3)( 1)

2(3)

sin 4 28

6

sin 4 28

6.215250437

sin 4 28

6 1.5485377

= 12.43º = 12º


Model Problem (Con’t)

In the interval 0º ≤ ≤ 360º, find to the nearestdegree, for all values of that satisfy the equation

3sin 4 1

sin

sin 4 28

6.215250437 = 12º

sine is (+) in QI & QII; reference = 12º

QI 12º QII 180 – 12 = 168ºand

{12º, 168º}


Aim: Solving Quadratic Trig Equations Course: Alg. 2 & Trig. Aim: How do we solve quadratic trigonometric equations? Do Now: Solve by factoring: x 2 –

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