標題:
f.4 maths
發問:
1 ) 2 sin2x -1 = 0 ( ans . x=15°,75°,195°,255° ) 2) 2 cos2x - √3 = 0 ( ans . x =15°,165°,195°,345° ) 3) solve 2 cosX = sin X +1 for ( 0°< X < 180°) ( ans . X = 36.9° )
最佳解答:
其他解答:
f.4 maths
發問:
1 ) 2 sin2x -1 = 0 ( ans . x=15°,75°,195°,255° ) 2) 2 cos2x - √3 = 0 ( ans . x =15°,165°,195°,345° ) 3) solve 2 cosX = sin X +1 for ( 0°< X < 180°) ( ans . X = 36.9° )
最佳解答:
此文章來自奇摩知識+如有不便請留言告知
2 sin2x -1=0 2sin 2x=1 sin 2x=1/2 0<180 so 0<2x<360 sin2x=sin 30 ,sin(180-30),sin(360+30),sin(360+180-30) 2x=30,21,390,550 x=15,75,195,255// 2 cos2x - √3 = 0 2cos 2x=√3 cos 2x=√3/2 cos 2x=cos 30 0<180 so 0<2x<360 cos 2x=cos 30,cos 360-30,cos (360+30),cos (360 +330) 2x=30,330,390,690 x=15,165,195,345 2cosX =sin X+1 4cos^2 x=sin^2 x+2sin x+1 4-4sin^2 x=sin^2 x+2sin x+1 5sin^2 x+2sin-3=0 (5sin x-3)(sin +1)=0 sinx=3/5 od sin=-1(reject,i.e.sin270=-1) sin x=sin 36.9 x=36.9//其他解答:
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