SOLUTION: Find, to the nearest degree, all values of x in the interval 0° ≤ x ≤ 360° that satisfy the equation 3 cos 2x + cos x + 2 = 0. I need your help please! I tried, but

Algebra ->  Trigonometry-basics -> SOLUTION: Find, to the nearest degree, all values of x in the interval 0° ≤ x ≤ 360° that satisfy the equation 3 cos 2x + cos x + 2 = 0. I need your help please! I tried, but       Log On


   

Question 389103: Find, to the nearest degree, all values of x in the interval 0° ≤ x ≤ 360° that satisfy the equation 3 cos 2x + cos x + 2 = 0.
I need your help please! I tried, but unable to come up with an answer.
Thanks
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Use the trig identity,
cos%282x%29=2%28cos%28x%29%29%5E2-1
So then,
3%2Acos%282x%29=6%28cos%28x%29%29%5E2-3
and then,
3%2Acos%282x%29+%2B+cos%28x%29+%2B+2+=+0
6%28cos%28x%29%29%5E2-3%2Bcos%28x%29%2B2=0
6%2A%28cos%28x%29%29%5E2%2Bcos%28x%29-1=0
Use a substitution,
Let u=cos%28x%29
6u%5E2%2Bu-1=0
You can factor this quadratic or solve it using the quadratic formula.
Once you have the solution for u, back substitute and find x.