Question

How do you solve `cos^2x+6cosx+4=0` in the interval [0,360]?

Answer 1

`arccos(-3+sqrt(5))`

Explanation:

Let `Y=cos x` then
`Y^2+6Y+4=0 and abs(Y)<=1`
Solving the equation
`Y_(1,2)=-3+-sqrt(9-4)=-3+-sqrt(5)`
Only the solution with + satisfy `abs(Y)>=1`

So

`cos x = -3+sqrt(5)`

and

`arccos(-3+sqrt(5))`