三角函數幾條問題

設所求θ之範圍為： 0 ≤ θ ≤ 360°


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1.
sinθ + cosθ = 0
sinθ = -cosθ
sinθ/cosθ = -1
tanθ = -1
θ = (180-45)°, (360-45)°
θ = 135°, 315°


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2.
2cos²θ + cosθ = 0
cosθ(2cosθ + 1) = 0
cosθ = 0 或 cosθ = -1/2
θ = 90°, 270° 或 θ = (180-60)°,(180+60)°
θ = 90°, 120°, 240°, 270°


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3.
5sin²θ - 2sinθcosθ - 3cos²θ = 0
(5sin²θ - 2sinθcosθ - 3cos²θ) / cos²θ = 0
5tan²θ - 2tan - 3 = 0
(5tanθ + 3)(tanθ - 1) = 0
tanθ = -3/5 或 tanθ = 1
θ = (180-30.96)°, (360-30.96)° 或θ = 45°, (180+45)°
θ = 45°, 149.04°, 225°, 329.04°


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4.
0 ≤ θ ≤ 360°
0 ≤ 3θ ≤ 1080°
-45° ≤ (3θ-45°) ≤ 1035°

cos(3θ-45°) = 1/2
3θ-45° = 60°, (360-60)°, (360+60)°, (720-60)°, (720+60)°, (1080-60)°
θ = 35°, 115°, 155°, 235°, 275°, 355°

.sinθ+cosθ=0
sinθ=-cosθ
tanθ=-1
θ=225°
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2cos^2θ+cosθ=0
cosθ(2cosθ+1)=0
cosθ=0 or -1/2
θ=90° 120° 240° or 270°
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5sin^2θ-2sinθcosθ-3cos^2θ=0
5tan^2 θ - 2tanθ - 3 = 0 [Whole line / cos^2 θ]
(tanθ-1)(5tanθ+3)=0
tanθ=1 or -3/5
θ=45° 149° 225° or 329°
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cos(3θ-45°)=1/2
3θ-45° = 60° or 300°
θ=35° or 115°