Precalc question

Question

Solve for x in

[0,2𝜋): 2cos𝑥−1−3sec𝑥=0

Answer 1

let y = cosx

2cosx -1 -3secx = 0

2cosx -1 - 3/cosx = 0

2y -1 - 3/y = 0

multiply by y

2y^2 -y -3= 0

factor

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

y =3/2, -1

cosx =3/2 is impossible since cosines are never greater than 1

cosx = -1

x = 180 degrees or pi radians

Answer 2

2cosx - 1 - 3secx = 0; 2cos²x - cosx - 3 = 0; (cosx + 1)(2cosx - 3) = 0; cosx + 1 = 0; cosx = -1, x = π

2cosx - 3 = 0, cosx = 3/2 > 1, no solutions,

Answer: x = π