6cos^2(x)-5cos(x)-6
just for readability, suppose u = cosx
Then you have
6u^2-5u-6=0
(3u+2)(2u-3)
that means
cosx = -2/3
cosx = 3/2
I guess you can take it from there, ok?
To solve the quadratic equation 6cos^2(x) - 5cos(x) - 6 = 0, we can use a substitution to simplify the equation. Let's use u = cos(x) to rewrite the equation in terms of u:
6u^2 - 5u - 6 = 0
To solve this quadratic equation, we can factor or use the quadratic formula:
Factorization method:
(3u + 2)(2u - 3) = 0
Setting each factor equal to zero:
3u + 2 = 0 or 2u - 3 = 0
Solving for u in each equation:
3u = -2 or 2u = 3
u = -2/3 or u = 3/2
Since u = cos(x), we can now solve for x:
1. For u = -2/3:
cos(x) = -2/3
To find x, we can use the inverse cosine function (arccos):
x = arccos(-2/3)
2. For u = 3/2:
cos(x) = 3/2
This equation has no real solutions since the range of cosine function is -1 to 1. Therefore, there are no solutions for this case.
So, the solution to the equation 6cos^2(x) - 5cos(x) - 6 = 0 is:
x = arccos(-2/3) + 2nπ, where n is an integer.