What are the solutions to the following ove the interval 0 to 360

Question

2 cos theta - 1 = sec theta

Answer 1

This one can be solved as a quadratic equation. Since sec θ = 1/cos θ, we have

 

2cos θ - 1 = sec θ  ⇒  2cos² θ - cos θ = 1

                            ⇒  2cos² θ - cos θ - 1 = 0,

 

which is quadratic in cos θ (if that isn't clear, try substituting x=cos θ above). This equation can be easily factorized, and doing so will give you two possible values for cos θ. At that point it should be easy to recognize which angles can give you those cosines; those angles are the solutions to the original equation.

Answer 2

Let x = theta

 

 

2cos(x) - 1 = sec(x)

 

 

Subtract sec(x) on both sides of equation.

 

2cos(x) - sec(x) - 1 = 0

 

 

Use the identity  sec(x) = 1 / cos(x)

 

2cos(x) - (1/cos(x)) - 1 = 0

 

 

The LCD is cos(x).

 

(2cos²(x) - 1 - cos(x)) / cos(x) = 0

 

 

Set the numerator equal to zero.

 

2cos²(x) - cos(x) - 1 = 0

 

(2cos(x) + 1)(cos(x) - 1) = 0

 

 

Set the factors equal to zero.

 

2cos(x) + 1 = 0               and            cos(x) - 1 = 0

 

2cos(x) = -1                       and           cos(x) = 1

 

cos(x) = -1/2                                           x = 0

                                                               x = 360

x = 60

x = 180 - 60 = 120

x = 180 + 60 = 240                              

 

 

 

The solutions are

 

x = 0

x = 120

x = 240

x = 360