Geometry help, please!

Question

Solve the following equations given that 0° ≤ θ < 360°:

 

2 cos 3θ = 1

 

 

Answer 1

cos(3θ) = 1/2

 

3θ = 60° + k360° or 300″ + k360°, k = 0, ±1. ±2, ...

 

θ = 20° + k120° or 100° + k120°

 

For k = 0, we get θ = 20°, 100°

 

For k = 1, we get θ = 140°, 220°

 

For k = 2, we get θ = 260°, 340°

 

Other choices for k give values of θ that lie outside of the interval  [0°, 360°)

 

Answer 2

2 cos 3θ = 1

cos 3θ = 1/2

3θ = π/3+2 π k, k=0,1,..

θ = π/9+(2/3)k π, 

if k=0, θ = π/9

if k=1, θ = π/9+2 π/3=7 π/9

if k=2, θ = π/9+4 π/3=13π/9

if k=3, θ = π/9+6 π/3=19 π/9> 2π 

So the solution(s): π/9, 7π/9, 12π/9

 

Answer 3

2 cos 3θ = 1

cos 3θ = 1/2

arccos(cos 3θ) = arccos(1/2)

3θ = 60^o, 300^o

θ = 20^o, 100^o