
William W. answered 11/11/20
Experienced Tutor and Retired Engineer
Let w = cos(x)
Then 2cos2(x) - 3cos(x) + 1 = 0 turns into 2w2 - 3w + 1 = 0 and you can factor that into (2w - 1)(w - 1) = 0 and solve giving you w = 1/2 or w = 1. Now back-substitute to get cos(x) = 1/2 or cos(x) = 1 and use the unit circle to solve. x = π/3, x = 5π/3, x = 0.
Since there are no domain restrictions you'd need to make the answers generic by adding 2πk to each:
x = π/3 + 2πk, x = 5π/3 + 2πk, x = 2πk