Question

how would you solve 2cos^2x + cosx - 1 = 0

Answer 1

This is a disguised quadratic equation. You need to make the substitution \[r = \cos(x)\] to get a quadratic equation in r. Then use the two values of \[r\] to solve the equation \[r=cos(x)\] for x.

Answer 2

so 2r^2 + r - 1 = 0?
I don't understand how that's solving it

Answer 3

r^2 + r - 2 (r+2)(r-1) .... (r+1) (2r-1).... (cosx + 1) (2cosx - 1).. then set them equal to 0?

Answer 4

Solve \[2r^2 + r - 2\] for r. Then we know that \[r = \cos(x),\] which we can use to solve for \[x.\]

Answer 5

\[2r^2 + r - 1 = 0\] actually. You can use the quadratic formula if you wish.

Answer 6

but was the way I did it right or? @jprahman?

Answer 7

the way you are doing it is right @jamroz

Answer 8

thank you! @TuringTest I was getting paranoid!