F4 maths question

- 3cos² θ - sinθcosθ + 1 = 0- 3cos² θ - sinθcosθ + (sin² θ + cos² θ) = 0sin² θ - sinθcosθ - 2cos² θ = 0(sin θ - 2 cos θ)(sin θ + cos θ) = 0sin θ - 2 cos θ = 0 or sin θ + cos θ = 0tan θ = 2 or tan θ = - 1θ = 63.43495°
or 180 + 63.43495 = 243.43495° for 0° =< θ <360°orθ = 135°
or 180 + 135 = 315° for 0° =< θ <360°


2011-01-04 20:34:53 補充：
Alternative :

- 3cos² θ - sinθcosθ + 1 = 0
- 3cos² θ - sinθcosθ + (sin² θ + cos² θ) = 0
sin² θ - sinθcosθ - 2cos² θ = 0
(sin² θ - sinθcosθ - 2cos² θ) / cos² θ = 0 / cos² θ
tan² θ - tan θ - 2 = 0
(tan θ - 2) (tan θ + 1) = 0
tan θ = 2 or tan θ = - 1

2011-01-04 20:34:59 補充：
θ = 180k + arctan 2 where k is integer.
or
θ = 180k + arctan - 1 = 180k + 3π/4 where k is integer.