中4maths

做D咩?!
解？
1. 2sin2θ+sinθ-1=0
(2sinθ-1)(sinθ+1)=0
sinθ=1/2 或 sinθ=-1
θ=30˚,150˚,270˚

2. tan2θ-2tanθ-3=0
(tanθ-3)(tanθ+1)=0
tanθ=3或tanθ=-1
θ=71.6˚,135˚,251.6˚,315˚

3. 4sin2θ+3sinθcosθ=2

1. 2sin^2θ+sinθ - 1 =0
(2sinθ-1)(sinθ+1)=0
sinθ=1/2=sin30 or sinθ=-1=-sin180
θ=30 or 180-30 or 180
θ=30 or 150 or 180

2 tan^2θ - 2tanθ-3 = 0
(tanθ-3)(tanθ+1)=0
tanθ=3=tan71.57 or tanθ=-1=-tan45
θ=71.6 or 251.6 or 135 or 315

3.4sin^2θ+3sinθcosθ =2
4sin^2θ+3sinθcosθ=2(sin^2θ+cos^2θ)
4sin^2θ+3sinθcosθ-2sin^2θ-2cos^2θ=0
2sin^2θ+3sinθcosθ-2cos^2θ=0
(2sinθ-cosθ)(sinθ+2cosθ)=0
2sinθ=cosθ or sinθ=-2cosθ
tanθ=1/2=tan26.6 or tanθ=-2=-tan63.4
θ=26.6 or 206.6 or θ=243.4 or 296.4

2008-03-04 19:18:48 補充：
sorry,
(1)sinθ=1/2=sin30 or sinθ=-1=sin270
θ=30 or 180-30 or 270
θ=30 or 150 or 270