Can I solve the equation like a quadratic? 2cos^2x + cos x-1=0

To solve for 2cos^2x+cosx-1 =
0

We rewite the equation
as:

2cos^2x + 2cosx -cosx -1 =
0.

We group th left
as:

(2cos^2x+2cosx) - (cosx+1) =
0

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

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

Equate the factors to
zero:

cosx+1 = 0

2cosx-1 =
0

cosx+1 = 0 gives cosx = 1, or x = pi/2 , 2npi+pi/2 0r x =
2npi-pi/2.

2cosx-1 = 0 gives cosx = 1/2. x = 2npi + pi/3 or
x = 2npi-pi/3.