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標題:
發問:
1.secθtanθ-tanθ+secθ-1=0 2.2cotθ-cscθ=0
最佳解答:
1. secθ tanθ - tanθ + secθ - 1 = 0 1/cosθ * sinθ/cosθ - sinθ/cosθ + 1/cosθ - 1 = 0 sinθ/cos2θ - sinθ/cosθ + 1/cosθ - 1 = 0 sinθ - sinθ cosθ + cosθ - cos2θ = 0 sinθ (1 - cosθ) + cosθ (1 - cosθ) = 0 (sinθ + cosθ)(1 - cosθ) = 0 sinθ = -cosθ or cosθ = 1 tanθ = -1 or cosθ = 1 θ = nπ - π/4 or θ = 2nπ, where n is integer【if you want general solution】 θ = 3π/4, 7π/4 or θ = 0, 2π【if 0≦θ≦2π】 θ = 135°, 315° or θ = 0°, 360°【if 0°≦θ≦360°】 2. 2cotθ - cscθ=0 2cosθ/sinθ - 1/sinθ = 0 (2cosθ - 1)/sinθ = 0 2cosθ - 1 = 0 cosθ = 1/2 θ = 2nπ ± π/3, where n is integer【if you want general solution】 θ = π/3, 5π/3【if 0≦θ≦2π】 θ = 60°, 300°【if 0°≦θ≦360°】 Hope it helps! ^^ 2006-12-09 14:07:26 補充： The key to do these questions is to change tan, sec, csc, cot into the basic form of sin and cos. Therefore, we can have more formula to work on them.
其他解答:1C924F1C0172E337
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解三角學方程發問:
1.secθtanθ-tanθ+secθ-1=0 2.2cotθ-cscθ=0
最佳解答:
1. secθ tanθ - tanθ + secθ - 1 = 0 1/cosθ * sinθ/cosθ - sinθ/cosθ + 1/cosθ - 1 = 0 sinθ/cos2θ - sinθ/cosθ + 1/cosθ - 1 = 0 sinθ - sinθ cosθ + cosθ - cos2θ = 0 sinθ (1 - cosθ) + cosθ (1 - cosθ) = 0 (sinθ + cosθ)(1 - cosθ) = 0 sinθ = -cosθ or cosθ = 1 tanθ = -1 or cosθ = 1 θ = nπ - π/4 or θ = 2nπ, where n is integer【if you want general solution】 θ = 3π/4, 7π/4 or θ = 0, 2π【if 0≦θ≦2π】 θ = 135°, 315° or θ = 0°, 360°【if 0°≦θ≦360°】 2. 2cotθ - cscθ=0 2cosθ/sinθ - 1/sinθ = 0 (2cosθ - 1)/sinθ = 0 2cosθ - 1 = 0 cosθ = 1/2 θ = 2nπ ± π/3, where n is integer【if you want general solution】 θ = π/3, 5π/3【if 0≦θ≦2π】 θ = 60°, 300°【if 0°≦θ≦360°】 Hope it helps! ^^ 2006-12-09 14:07:26 補充： The key to do these questions is to change tan, sec, csc, cot into the basic form of sin and cos. Therefore, we can have more formula to work on them.
其他解答:1C924F1C0172E337
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