Solve the equation for the given domain

Question

1.) cos²Θ + cosΘ-1=0 for 0°<=Θ<=180°

 

2.) 4cos²Θ=1 for 0°<=Θ<=360°

Answer 1

1)  cos^2θ+cosθ-1=0

     complete the square

     cos^2θ+cosθ   =1

     [(1/2)(1)]^2=1/4

     cos^2θ+cosθ+(1/4)=1+(1/4)

     (cosθ+[1/2])^2=1.25

     (cosθ+[1/2])=√1.25

     cosθ+[1/2]=1.118033989

     cosθ=1.118033989-0.5

     cosθ=0.618033988

     use arcos finder on your calculator

     θ=51.82729243º or 51º 49' 38.253"

    

 

Answer 2

Hi Hannah,

 

2) 4cos^2theta=1

cos^2 theta = 1/4

cos theta = +/- 1/2

theta = 60°, 120°, 240° and 300°