solve 2cos^2x + cosx - 1 = 0, i think the trig part of my brain was taken out without my knowing...?

Question

i knew how to do this last week :/

Answer 1

First, note that this is a quadratic equation in cos x, so use the quadratic formula:

cos x = (-1 ± √(1 + 8))/4 = (-1±3)/4
cos x = -4/4 = -1 or cos x = 2/4 = 1/2

So we have cos x = -1 or cos x = 1/2. The values of x for which cos x = -1 are π + 2πk for every integer k, and the values of x for which cos x = 1/2 are π/3 + 2πk and -π/3 + 2πk for every integer k. So the solution set is:

{π + 2πk: k∈Z} ∪ {π/3 + 2πk: k∈Z} ∪ {-π/3 + 2πk: k∈Z}

And we are done.

Answer 2

Do a substitution with W = cos x.
The equation becomes 2W^2 + W - 1 = 0 which is just a standard quadratic equation.
Solve to get W = 0.5 or -1

For W = cos x = 0.5, you get x = pi/3 or -pi/3 (60 or -60 degrees)
For W = cos x = -1, you get x = pi (180 degrees)

Answer 3

You gotta use the algebra part of your brain first ; ) !

cos x = u

2u^2 + u - 1= 0
2u^2 + 2u - u - 1 = 0
2u(u+1) - (u+1) = 0
(2u - 1)(u + 1) = 0

cos x = -1
cos x = 1/2

Think you can find x now?