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

Question

anonymous · Answer

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.

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

turingtest · Answer

the way you are doing it is right @jamroz

anonymous · Answer

thank you! @TuringTest I was getting paranoid!