一些三角恒等式的問題(急)

tanθ =2sinθ
sin θ / cos θ = 2 sin θ
sin θ = 2 sin θ cos θ
2 sin θ cos θ - sin θ = 0
sin θ (2 cos θ - 1) = 0
sin θ = 0 or cos θ = 1/2

a) 0° ≦θ≦ 90°
sin θ = 0 => θ = 0°
cos θ = 1/2 => θ = 60°
So θ = 0° or 60°.

b) 0° ≦θ≦ 360°
sin θ = 0 => θ = 0°, 180°, 360°
cos θ = 1/2 => θ = 60°, 300°
So θ = 0°, 60°, 180°, 300° or 360°.

2. tan² θ + 2 tan θ - 3 = 0
(tan θ - 1)(tan θ + 3) = 0
tan θ = 1 or tan θ = -3

a) 0° ≦θ≦ 90°
tan θ = 1 => θ = 45°
tan θ = -3 => There is no solution for θ
So θ = 45°.

b) 0° ≦θ≦ 360°
tan θ = 1 => θ = 45°, 225°
tan θ = -3 => θ = 108.4° or 288.4°
So θ = 45°, 108.4°, 225° or 288.4°.

2006-12-20 22:32:33 補充：
不好意思, 以上兩位朋友第一題都答錯. 這些都是常見的.sinθ =sin2θ, 不能直接推斷為 θ =2θ.tanθ=sinθ/cosθ=2sinθ => 1/cosθ=2也是錯, 除非題目寫明sin θ不等於零. 若果等於零, 這様兩邊除以sin θ就等於除以0, 變成undefined的情況.

1a)
tanθ =2sinθ
tanθ=sinθ/cosθ=2sinθ
1/cosθ=2
1/2=cosθ
θ=60
1b)θ=60,300
2a)
(tanθ)^2+2tanθ-3=0
let tanθ=x
x^2+2x-3=0
(x+3)(x-1)=0
x=-3=tanθ
x=1=tanθ
θ=45
2b)θ=45,108.435,288.435,225