Download Algebra III Lesson 60

Algebra III
Lesson 60
Factorable Trigonometric Equations –
Loss of Solutions Caused by Division
Factorable Trigonometric Equations
Factor:
x2 – 1 = 0
sin2 x – 1 = 0
(x – 1)(x + 1) = 0
(sin x – 1)(sin x + 1) = 0
x2 + x = 0
tan2 θ + tan θ = 0
x(x + 1) = 0
tan θ(tan θ + 1) = 0
Example 60.1
Solve tan2 θ – 1 = 0 given that 0° ≤ θ < 360°.
(tan θ – 1)(tan θ + 1) = 0
tan θ – 1 = 0
tan θ + 1 = 0
tan θ = 1
tan θ = -1
Remember tan = y/x and use the unit circle.
θ = 45°, 135°, 225°, 315°
Example 60.2
Solve sin2 θ – sin θ = 0 given that 0° ≤ θ < 360°.
sin θ(sin θ - 1) = 0
sin θ = 0
sin θ – 1 = 0
sin θ = 1
θ = 0°, 90°, 180°
Example 60.3
Solve cos2 θ = 1 given that 0° ≤ θ < 360°.
cos2 θ – 1 = 0
(cos θ – 1)(cos θ + 1) = 0
cos θ – 1 = 0
cos θ = 1
θ = 0°, 180°
cos θ + 1 = 0
cos θ = -1
Loss of Solutions Caused by Division
Consider
2sin x cos x = sin x
Tempting to just divide by sin x and get.
2cos x = 1
cos x = ½
x = 60°, 300°
Now look at it another way.
Example 60.4
Solve 2sin x cos x = sin x given that 0° ≤ θ < 360°.
2sinx cos x – sin x = 0
sin x(2cos x – 1) = 0
sin x = 0
2 cos x – 1 = 0
cos x = ½
x = 0°, 60°, 180°, 300°
Practice
a) Solve cos2 θ – cos θ = 0 given that 0° ≤ θ < 360°.
cos θ(cos θ - 1) = 0
cos θ – 1 = 0
cos θ = 1
cos θ = 0
θ = 0°, 90°, 270°
b) Simplify:
log 3 6 − log 3 2
3
log 3 6 / 2
3
=3
c) How many distinguishable ways can a host and 10 guests be seated
in a row if three of the guests are identical triplets?
P =
n!
a!
=
11!
11⋅10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3!
=
3!
3!
= 6652800
d) Complete the square and graph the parabola y = 6x + 3x2.
y = 3(x2 + 2x
)
y = 3(x2 + 2x + 1) - 3
y = 3(x + 1)2 – 3