I need exact values between 0<- x >2pi
2006-11-21 03:32:38 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
2cos(x) - sin(2x) = 0
remembering that sin(2x) = 2sin(x)cos(x),
2cos(x) -2sin(x)cos(x) = 0
2cos(x) * [1 - sin(x)] = 0
or cos(x) = 0 or 1-sin(x) = 0
cos(x) = 0 or sin(x) = 1
π/2, 3π/2
2006-11-21 03:37:13 · answer #1 · answered by Scott R 6 · 3⤊ 2⤋
2 cosx - 2. sin x . cos x = 0
2 cos x( 1- sinx) = 0
2 cos x = 0 and 1- sin x = 0
x = pi/2 or 3pi/2 sin x = 1
x = pi/ 2
2006-11-21 03:41:45 · answer #2 · answered by bmonica01 1 · 1⤊ 3⤋
sin2x=2sinxcosx
2cosx-sinxcosx=0
2cosx(1-sinx)=0
then , cosx=0 x=pi/2
& sinx=1 x =3pi/2
2006-11-21 03:42:32 · answer #3 · answered by KURTANGLE 4 · 2⤊ 2⤋
2cosx-sin2x=0
2cosx-2cosx*sinx=0
cosx[1-sinx]=0
sinx=0
x=0 deg
cosx=0
x=90Deg
2006-11-21 03:45:08 · answer #4 · answered by openpsychy 6 · 2⤊ 3⤋