Solve: (1-√3tanθ)+1+√3 = √3sec^2θ Where (0≤θ≤360)

Answer 1

recall that sec^2 Ø = tan^2 Ø + 1

(1-√3tanθ)+1+√3 = √3sec^2θ
1 - √3tanØ + 1 + √3 = √3(tan^2 Ø + 1)
- √3tanØ + √3 = √3tan^2 Ø + √3

tan^2 Ø + tanØ = 0
tanØ(tanØ + 1) = 0
tanØ = 0 or tanØ = -1
Ø = 0 or Ø = 135° or Ø = 315°

Answer 2

in the solution of tanØ = 0
I should have included Ø = 180°

so Ø = 0°, 135°, 180°, 315°