sin 5x + sin X = 2[sin 3x + cos 2X] (from compound angle formula) how do i take it from here???
2007-05-03 23:45:43 · 6 answers · asked by torpedo 1 in Science & Mathematics ➔ Mathematics
sorry, i have to find the solutions for this equattion,, between o and pi,,
could i go from here by saying that [sin 5x + sin X] = s[sin 3x + cos 2X]
or show me any other way of doing this
2007-05-03 23:58:14 · update #1
sin 5x + sin x=0
2 sin(6x/2)cos(4x/2)=0
sin3x.cos 2x=0
sin3x=0
3x=0 or 3x=pi
x1=0
x2=pi/3
or
cos 2x=0
2x=pi/2
x3=pi/4
2x=3pi/2
x4=3pi/4
Solution set={0,pi/3,pi/4, 3pi/4}
2007-05-04 00:33:28 · answer #1 · answered by iyiogrenci 6 · 0⤊ 0⤋
To SOLVE an equation an = sign is required.
All that can be done is to write sin 5x + sin x in another form.
sin 5x + sin x = 2.sin 3x.cos 2x
2007-05-04 00:56:41 · answer #2 · answered by Como 7 · 0⤊ 0⤋
use formula:
sin a + sin b = 2 sin ((a+b)/2) cos ((a-b)/2)
2 sin 3x cos 2x
2007-05-04 00:11:28 · answer #3 · answered by nelaq 4 · 0⤊ 0⤋
sin 5x + sin X = 2[sin 3x + cos 2X]
2sin3x*cos2x=2
2007-05-03 23:56:05 · answer #4 · answered by Anonymous · 0⤊ 0⤋
What are you solving for? Trying to get it just in terms of sin x and cos x? Or is it equal to something?
We can't solve the equation for X unless you set it equal to something.
2007-05-03 23:56:24 · answer #5 · answered by Mathematica 7 · 0⤊ 0⤋
sin 5x sin 3x / sin 3x + sin x
2015-08-11 13:45:06 · answer #6 · answered by ? 1 · 0⤊ 0⤋