Exact Solution

Question

Find the EXACT solutions of the given equation in the interval [0,2π). There may be more than one. 

6cos2(x) + 5cos(x) − 4 = 0

Answer 1

6cos(2x) + 5cos(x) - 4 = 0

 

6(cos²x - sin²x) + 5cos(x) - 4 = 4

 

 

Let's get everything in terms of cosine.

 

 

6(cos²x - (1 - cos²x)) + 5cos(x) - 4 = 0

 

6(cos²x - 1 + cos²x) + 5cos(x) - 4 = 0

 

6(2cos²(x) - 1) + 5cos(x) - 4 = 0

 

12cos²(x) - 6 + 5cos(x) - 4 = 0

 

12cos²(x) + 5cos(x) - 10 = 0

.

We can treat this like a quadratic equation. We let x = cos(x).

 

cos(x) = (-5 ± √(25 - 4(-120))) / 24

          = (-5 ± √(505)) / 24

          = (-5 ± 22.47) / 24

 

cos(x) = -1.14           and          cos(x) = 0.73

 

x = cos^-1(-1.14)          and             x = cos^-1(0.73)

 

                                                    x = 43.11

 

Cosine is positive in the first and fourth quadrant.  We can find the other solution by subtracting this value of x by 360.

 

x = 360 - 43.11

x = 316.89

 

We have to solutions:

 

x = 43.11        and        x = 316.89

 

 These solutions are in degrees.