How to solve 8 cos (x)^3-6 cos (x)-1=0

Answer 1

Answer 2

Hi Ashton,

Do you remember the trig identity cos³x=1/4(3cos + cos3x)? If we multiply this by 8 we can replace the first term with 6cosx + 2cos3x and that makes the given equation 2cos3x=1. Dividing by 2 gives us the equation cos3x=1/2 now solving for x gives x=π/9 which is one solution. I think there are others. See if you can find them.

Regards

Jim

Answer 3

Let y = cos x

8y³ - 6y - 1 = 0 has 3 real irrational roots which can be found using your graphing calculator (or other root finding method).

Then the roots of the original equation are arccos of each of those roots. Please note that there will be multiple values for each root of the quadratic in y since cos is periodic.