@ represents theta
solve for @ in terms of radians, not degrees
2007-05-16 21:57:44 · 4 answers · asked by markdosdourian 1 in Science & Mathematics ➔ Mathematics
2sin@cos@ = sin@
DO NOT CANCEL THE SIN@
2sin@cos@ - sin@ = 0
sin@ * (2cos@ - 1) = 0
sin@ = 0
@ = 0, pi, 2pi
cos@ = 1/2
@ = pi/3, 5*pi/3
2007-05-16 22:00:43 · answer #1 · answered by Dr D 7 · 2⤊ 1⤋
I will just call it x
sin2x = sinx
2sinxcosx=sinx
2sinxcosx-sinx=0
sinx(2cosx-1)=0
sinx=0
x=0, pi, 2pi
2cosx=1
cosx=1/2
x=pi/3, 5pi/3
Since 0
Solutions are pi/3, pi and 5pi/3
2007-05-17 05:04:31 · answer #2 · answered by gudspeling 7 · 0⤊ 0⤋
Let sin 2x = sin x
2.sin x.cos x = sin x
sin x.(2 cos x - 1) = 0
sin x = 0 , cos x = 1/2
x = 0 , 2Ï , Ï/3 , 5Ï/3
x = Ï/3 , 5Ï/3 (to satisfy limits given for x)
2007-05-17 05:08:15 · answer #3 · answered by Como 7 · 0⤊ 0⤋
2sin@cos@ = sin@
2sin@cos@ - sin@ = 0
sin@ * (2cos@ - 1) = 0
sin@ = 0
@ = 0, pi, 2pi
or
2cos@ - 1 = 0
cos@ = 1/2
@ = pi/3, 5*pi/3
@ = 0, pi, 2pi, pi/3, 5*pi/3
2007-05-17 05:18:23 · answer #4 · answered by Kinu Sharma 2 · 0⤊ 1⤋