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功課一問......三角方程

發問:

解下列各方程,其中0°大於θ大於360°。 1. cosθ(tanθ+1)=0 2. (3cosθ+2)(sinθ-2)=0 3. 4tan2次θ-11tanθ+7=0 4. 3sin2次θ-2cosθ-2=0

最佳解答:

Q1. cosθ(tanθ+1)=0 A: cosθ(tanθ+1)=0 cosθ= 0 or (tanθ+1)=0 θ= 90° or 270° or tanθ= -1 θ= 135° or 315° Q2. (3cosθ+2)(sinθ-2)=0 A: (3cosθ+2)(sinθ-2)=0 (3cosθ+2) = 0 or (sinθ-2)=0 cosθ= -2/3 or sinθ= 2 (rejected) θ= 131.81° or 228.19° Q3. 4tan2次θ-11tanθ+7=0 A: 4tan2次θ-11tanθ+7=0 (4tanθ-7)(tanθ-1)=0 (4tanθ-7)=0 or (tanθ-1)=0 tanθ=7/4 or tanθ=1 θ= 60.26° or 240.26° or θ= 45° or 225° Q4. 3sin2次θ-2cosθ-2=0 A: 3sin2次θ-2cosθ-2=0 3(1-cos2次θ)-2cosθ-2=0 3-3cos2次θ-2cosθ-2=0 3cos2次θ+2cosθ-1 =0 (3cosθ-1)(cosθ+1) = 0 (3cosθ-1)=0 or (cosθ+1) = 0 cosθ=1/3 or cosθ=-1 θ= 70.59° or 289.41° or θ= 180° For your reference. Please rearrange each answers in order (=你可在最後把答案順序才回答).

其他解答:4E350C6F8B48ECA2
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