2017-02-14 14:24:37dvhtlbz
F4 maths question
標題:
F4 maths question
-3cos^2 Θ - sinΘcosΘ + 1 = 0
最佳解答:
- 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.
其他解答:A215E4A27F624C16
F4 maths question
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發問:-3cos^2 Θ - sinΘcosΘ + 1 = 0
最佳解答:
- 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.
其他解答:A215E4A27F624C16