Verified answer

This is what is known as a special quadratic

2cos(x)^2 - cos(x) = 3

2cos(x)^2 - cos(x) - 3 = 0

cos(x) = (1 +/- sqrt(1 - 4 * 2 * (-3))) / (2 * 2)

cos(x) = (1 +/- sqrt(1 + 24)) / 4

cos(x) = (1 +/- sqrt(25)) / 4

cos(x) = (1 +/- 5) / 4

cos(x) = 6/4 , -4/4

cos(x) = 3/2 , -1

cos(x) can never equal 3/2

cos(x) = -1

x = pi

Cos 3 0

Rearrange into a quadratic equation and factorise

2cosx^2 - cosx = 3

2cosx^2 - cosx - 3 = 0

(2cosx - 3)(cosx + 1) = 0

cosx = 3/2 or cos x = - 1 It can't = 3/2

So cos x = - 1

x = pi.

factor.

2 cos^2 x - cos x - 3 = 0

(2 cos x - 3)(cos x + 1) = 0

2 cos x - 3 = 0

2 cos x = 3

cos x = 3/2

cos x + 1 = 0

cos x = -1

subtract 3 from both sides.

2cos^2x-cosx-3=0

let u=cosx

2(cosx)^2-cosx-3=0

2u^2-u-3=0

(2u-3)(u+1)=0

u=3/2 u=-1

cosx=3/2 cosx=-1

cosx can't ever by 3/2 because when you take the arccos of both sides you get x=cos^-1(3/2) and that isn't possible.

cosx=-1

x=pi.

general solution:

x=2npi+pi

now insert values for n

n=0 x=pi

n=1 x=2pi

I don't know if you have some time of restriction but most people only accept answer between 0 and 2pi.

Solutions are: x=pi.